The Exponential Function

If $x=0$, the series is $1+0+0+\cdots$ and converges absolutely. Hence assume $x\mathrm{\ne }0$. Choose $N\ge 1$ with $\iota (N)>2\mathrm{\mid }x\mathrm{\mid }$. For $n\ge N$, the absolute terms $a_n=\mathrm{\mid }x{\mathrm{\mid }}^n\mathrm{/}\iota (n!)$ are positive and satisfy $a_{n+1}\mathrm{/}{a}_n=\mathrm{\mid }x\mathrm{\mid }\mathrm{/}\iota (n+1)<1\mathrm{/}2$.

Thus $a_{N+j}\le a_N{2}^{-j}$ by induction, and the tail is dominated by a convergent geometric series.

The zero case from step 1.1 and, when $x\mathrm{\ne }0$, adding the finite initial segment to the convergent tail prove absolute convergence for arbitrary $x$. Hence every nonnegative radius works and the radius is $+\mathrm{\infty }$.

Apply [L1] at the auxiliary value $z=1$. The coefficient of degree $n$ in the resulting Cauchy product for $\exp (x)\exp (y)$ is ${\sum }_{k=0}^n{x}^k{y}^{n-k}\mathrm{/}(\iota (k!)\iota ((n-k)!))$.

Apply [L3] and [L2] to identify this finite sum with $(x+y{)}^n\mathrm{/}\iota (n!)$.

Summing over $n$ gives the exponential series at $x+y$, hence the formula.

Setting $y=-x$ in [L1] gives $\exp (x)\exp (-x)=1$, so both factors are nonzero.

Also $\exp (x)=\exp (x\mathrm{/}2{)}^2$, so it is nonnegative; by step 1.1 and [L2] it is positive.

Dividing the identity in step 1.1 by $\exp (x)$ gives the reciprocal formula.

Termwise differentiation gives $(\exp {)}^{\mathrm{\prime }}(x)={\sum }_{n\ge 1}\iota (n)x^{n-1}\mathrm{/}\iota (n!)$.

Reindex $n=j+1$ and cancel $\iota (j+1)$ using the factorial recurrence. The series becomes ${\sum }_{j\ge 0}{x}^j\mathrm{/}\iota (j!)=\exp (x)$.

Smoothness follows from [L1] and the infinite radius; iterating step 2.1 gives every higher derivative.

If $x<y$, the mean value theorem gives $\exp (y)-\exp (x)=\exp (c)(y-x)$ for some $c\in (x,y)$.

Both factors on the right are positive, so $\exp (y)>\exp (x)$. Continuity is the cited power-series conclusion.

Given a real $M$, every $x>\max \{0,M-1\}$ satisfies $\exp (x)\ge 1+x>M$. Hence $\exp (x)\to +\mathrm{\infty }$.

Given $\epsilon >0$, choose $X>0$ with $1+X>1\mathrm{/}\epsilon$. If $x<-X$, then $-x>X$, so [L1] gives $\exp (-x)\ge 1-x>1+X>1\mathrm{/}\epsilon$; [L2] yields $0<\exp (x)<\epsilon$.

The range assertions follow from positivity and the two limit conclusions.

Strict increase gives injectivity.

Given $y>0$, [L2] provides $a<b$ with $\exp (a)<y<\exp (b)$. Applying [L3] on $[a,b]$ gives $c$ with $\exp (c)=y$.

Positivity gives the stated codomain, and steps 1.1 and 1.2 give bijectivity.

For $x>0$, retain term $m+1$ of the series at $ax$: $\exp (ax)\ge (ax{)}^{m+1}\mathrm{/}\iota ((m+1)!)$.

Hence $0\le x^m\mathrm{/}\exp (ax)\le \iota ((m+1)!)\mathrm{/}(a^{m+1}x)$.

The upper bound tends to $0$, so the quotient tends to $0$.

Define $h(x)=y(x)\exp (-x)$. By [L1] and [L2], $h^{\mathrm{\prime }}(x)=y^{\mathrm{\prime }}(x)\exp (-x)-y(x)\exp (-x)=0$.

The differentiable function $h$ is continuous, so [L3] makes it constant; $h(0)=y(0)\exp (0)=1$.

Thus $y(x)\exp (-x)=1$, and multiplying by $\exp (x)$ gives $y(x)=\exp (x)$.

Under continuity and $F(1)=e$, the equation gives $F(n)=e^n$, $F(-n)=e^{-n}$, and uniqueness of positive roots gives $F(m\mathrm{/}n)=e^{m\mathrm{/}n}$ for rationals $m\mathrm{/}n$. Density and continuity then give $F(x)=\exp (x)$ for every real $x$.

Under differentiability at $0$, $\frac{F(x+h)-F(x)}{h}=F(x)\frac{F(h)-1}{h}$, so $F^{\mathrm{\prime }}(x)=F(x)F^{\mathrm{\prime }}(0)=F(x)$. With $F(0)=1$, [L1] gives $F=\exp$.

The exponential itself satisfies both normalizations, so both uniqueness assertions follow.

For $n\ge \max \{1,k\}$, $\iota \left(\genfrac{}{}{0px}{}{n}{k}\right)\mathrm{/}\iota (n{)}^k=\iota (k!{)}^{-1}{\prod }_{j<k}(1-\iota (j)\mathrm{/}\iota (n))$.

For $j<k\le n$, strict increase and positivity give $0\le \iota (j)\mathrm{/}\iota (n)<1$, so every factor in step 1.1 lies in $(0,1]$. Thus the finite product lies in $[0,1]$, proving the displayed uniform bound.

For each of the finitely many $j<k$, $\iota (j)\mathrm{/}\iota (n)\to 0$; finite-product limit algebra makes the product in step 1.1 tend to $1$. Multiplication by the fixed factor $1\mathrm{/}\iota (k!)$ yields the limit.

For $n>\mathrm{\mid }x\mathrm{\mid }$, the binomial theorem gives $(1+x\mathrm{/}\iota (n){)}^n={\sum }_{k=0}^n(\iota \left(\genfrac{}{}{0px}{}{n}{k}\right)\mathrm{/}\iota (n{)}^k)x^k$.

Each fixed coefficient tends to $1\mathrm{/}\iota (k!)$, while the uniform bound in [L1] holds for every term present in the sum.

Given $\epsilon >0$, choose $K$ so the absolute exponential tail after $K$ is below $\epsilon \mathrm{/}3$ using [L2]. The same coefficient bound controls the product tail uniformly in $n$; for the finite head $k\le K$, choose $n$ so all coefficient errors sum to below $\epsilon \mathrm{/}3$.

The triangle inequality then makes the product differ from ${\sum }_{k\ge 0}{x}^k\mathrm{/}\iota (k!)=\exp (x)$ by less than $\epsilon$.

At $r=0$, $u_0=1$, the stated finite sum.

If the formula holds at $r$, integrate its finite sum termwise from $0$ to $x$. By [L1], the integral of $t^k\mathrm{/}\iota (k!)$ is $x^{k+1}\mathrm{/}\iota ((k+1)!)$, giving the formula at $r+1$.

Hence the iterates are precisely the partial sums of the exponential series. Its infinite radius and [L2] give uniform convergence on every bounded interval.

Fix $x$ and work on the compact interval with endpoints $0$ and $x$. The polynomial iterates are continuous and integrable there, and step 2.1 gives uniform convergence to $\exp$. Thus [L3] lets the integrals in $u_{r+1}(x)=1+{\int }_0^x{u}_r(t)dt$ pass to the limit, giving $\exp (x)=1+{\int }_0^x\exp (t)dt$, with the orientation supplied by The integral with oriented limits: ${\int }_a^{a}f:=0$ and ${\int }_b^{a}f:=-{\int }_a^{b}f$ when $x<0$.

Step 2.1 and [L3] make $\exp$ continuous. The first fundamental theorem applied to step 3.1 gives ${\exp }^{\mathrm{\prime }}(x)=\exp (x)$, and setting $x=0$ gives $\exp (0)=1$.

Each theorem in [L2] identifies its construction with the series-defined function in [L1], with exactly the normalization stated here.

Equality with a common function is transitive, so all five descriptions are equivalent.

For $k\ge N+1$, strict increase gives $\iota (k+1)\ge \iota (N+2)\ge 2\mathrm{\mid }x\mathrm{\mid }$, and the factorial recurrence gives that the ratio of consecutive absolute terms is $\mathrm{\mid }x\mathrm{\mid }\mathrm{/}\iota (k+1)\le 1\mathrm{/}2$.

Thus the $j$-th term after $N+1$ is at most the first tail term times $2^{-j}$.

Sum the geometric majorant using [L2] to obtain the displayed bound.

The first three terms give $e>1+1+1\mathrm{/}2>2$.

The term at index $2$ is $1\mathrm{/}2$. Apply [L1] with $x=1,N=2$: the tail from index $3$ onward is at most $2\mathrm{/}3!=1\mathrm{/}3$. Thus the whole tail after index $1$ is at most $1\mathrm{/}2+1\mathrm{/}3=5\mathrm{/}6<1$, and hence $e<2+1=3$.

Combining the strict bounds gives the claim.

Assume $e\in \mathbb{Q}$. By [L3], write $e=p\mathrm{/}q$ in $\mathbb{R}$ with $p\in \mathbb{Z}$ and $q\in \mathbb{N}$, $q\ge 1$, using the canonical embeddings. Choose a natural $n\ge \max \{q,2\}$ (Every complete ordered field is Archimedean).

Put $A:=\iota (n!)(e-{\sum }_{k=0}^{n}1\mathrm{/}\iota (k!))$. Every tail term is positive, so $A>0$. Applying [L2] with $x=1$ and $N=n$, then using the factorial recurrence, gives $A\le \frac{2\iota (n!)}{\iota ((n+1)!)}=\frac{2}{\iota (n+1)}\le \frac{2}{3}<1$ because $n\ge 2$.

The number $A$ from step 2.1 is an embedded integer. Indeed, for each $0\le k\le n$, [L1] gives a natural $s_k$ with $n!=k!s_k$. Also $q!=m!q$ for the natural $m$ with $q=m+1$, and [L1] at $k=q$ gives $q!\mid n!$; hence $n!=qr$ for some natural $r$. By [L3] and multiplicativity of the embeddings, $\iota (n!)e=\widehat{pr},\frac{\iota (n!)}{\iota (k!)}=\iota (s_k),$ where $\widehat{pr}$ is the real image of the integer $pr$. Therefore $A$ is a difference of embedded integers and is itself an embedded integer.

Since the embedding preserves order, no embedded integer lies strictly between $0$ and $1$, contradicting steps 3.1 and 2.1. Therefore $e\mathrm{\notin }\mathbb{Q}$.