The Exponential Function

If $x=0$, the series is $1+0+0+\cdots$ and converges absolutely. Hence assume $x\ne0$. Choose $N\ge1$ with $\iota(N)>2|x|$. For $n\ge N$, the absolute terms $a_n=|x|^n/\iota(n!)$ are positive and satisfy $a_{n+1}/a_n=|x|/\iota(n+1)<1/2$.

Thus $a_{N+j}\le a_N2^{-j}$ by induction, and the tail is dominated by a convergent geometric series.

The zero case from step 1.1 and, when $x\ne0$, adding the finite initial segment to the convergent tail prove absolute convergence for arbitrary $x$. Hence every nonnegative radius works and the radius is $+\infty$.

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

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

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 $\exp(x)=\exp(x/2)^2$, 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 $(\exp)'(x)=\sum_{n\ge1}\iota(n)x^{n-1}/\iota(n!)$.

Reindex $n=j+1$ and cancel $\iota(j+1)$ using the factorial recurrence. The series becomes $\sum_{j\ge0}x^j/\iota(j!)=\exp(x)$.

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 $\exp(x)\ge1+x>M$. Hence $\exp(x)\to+\infty$.

Given $\varepsilon>0$, choose $X>0$ with $1+X>1/\varepsilon$. If $x<-X$, then $-x>X$, so [L1] gives $\exp(-x)\ge1-x>1+X>1/\varepsilon$; [L2] yields $0<\exp(x)<\varepsilon$.

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 $ax$: $\exp(ax)\ge (ax)^{m+1}/\iota((m+1)!)$.

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

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

Define $h(x)=y(x)\exp(-x)$. By [L1] and [L2], $h'(x)=y'(x)\exp(-x)-y(x)\exp(-x)=0$.

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 $F(m/n)=e^{m/n}$ for rationals $m/n$. 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 $F'(x)=F(x)F'(0)=F(x)$. With $F(0)=1$, [L1] gives $F=\exp$.

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

For $n\ge\max\{1,k\}$, $\iota\binom nk/\iota(n)^k=\iota(k!)^{-1}\prod_{j<k}(1-\iota(j)/\iota(n))$.

For $j<k\le n$, strict increase and positivity give $0\le\iota(j)/\iota(n)<1$, 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 $j<k$, $\iota(j)/\iota(n)\to0$; finite-product limit algebra makes the product in step 1.1 tend to $1$. Multiplication by the fixed factor $1/\iota(k!)$ yields the limit.

For $n>|x|$, the binomial theorem gives $(1+x/\iota(n))^n=\sum_{k=0}^{n}\bigl(\iota\binom nk/\iota(n)^k\bigr)x^k$.

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

Given $\varepsilon>0$, choose $K$ so the absolute exponential tail after $K$ is below $\varepsilon/3$ 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 $\varepsilon/3$.

The triangle inequality then makes the product differ from $\sum_{k\ge0}x^k/\iota(k!)=\exp(x)$ by less than $\varepsilon$.

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 $t^k/\iota(k!)$ is $x^{k+1}/\iota((k+1)!)$, 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 $\exp'(x)=\exp(x)$, 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 $\iota(k+1)\ge\iota(N+2)\ge2|x|$, and the factorial recurrence gives that the ratio of consecutive absolute terms is $|x|/\iota(k+1)\le1/2$.

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 $e>1+1+1/2>2$.

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

Combining the strict bounds gives the claim.

Assume $e\in\mathbb Q$. By [L3], write $e=p/q$ in $\mathbb R$ with $p\in\mathbb Z$ and $q\in\mathbb N$, $q\ge1$, using the canonical embeddings. Choose a natural $n\ge\max\{q,2\}$ (Every complete ordered field is Archimedean).

Put $A:=\iota(n!)\left(e-\sum_{k=0}^{n}1/\iota(k!)\right)$. Every tail term is positive, so $A>0$. Applying [L2] with $x=1$ and $N=n$, then using the factorial recurrence, gives $A\le \frac{2\iota(n!)}{\iota((n+1)!)} =\frac2{\iota(n+1)} \le\frac23<1$ because $n\ge2$.

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, $\iota(n!)e=\widehat{pr},\qquad \frac{\iota(n!)}{\iota(k!)}=\iota(s_k),$ 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 $e\notin\mathbb Q$.