The exponential addition formula $\exp(x+y)=\exp(x)\exp(y)$

Given: $x,y\in\mathbb R$.

For fixed $x,y$, the auxiliary power series $\sum_{n\ge0}x^nz^n/\iota(n!)$ and $\sum_{n\ge0}y^nz^n/\iota(n!)$ have infinite radius by The exponential series converges absolutely for every real argument. Inside their common radius, their product is the Cauchy product of their coefficients (Inside the common radius the product of two power-series sums is represented by the Cauchy product of their coefficients).

The binomial theorem says $(x+y)^n=\sum_{k=0}^n\iota\binom nk x^ky^{n-k}$ (The binomial theorem in $\mathbb{R}$: $(x+y)^{n} = \sum_{k<n+1} \iota\!\binom{n}{k}\, x^{k} y^{\,n-k}$, The set $[A]^{k}$ of $k$-element subsets and the binomial coefficient $\binom{n}{k} := \lvert [n]^{k}\rvert$).

For $k\le n$, $\iota\binom nk=\iota(n!)/(\iota(k!)\iota((n-k)!))$ ($\binom{n}{k}\,k!\,(n-k)! = n!$ for $k \le n$; hence $\binom{n}{k}\,k! = n^{\underline{k}}$, the quotient $n!/(k!(n-k)!)$ is a natural number, and $\binom{n}{k} = \binom{n}{n-k}$). Therefore $1/(\iota(k!)\iota((n-k)!))=\iota\binom nk/\iota(n!)$, with all naturals read in $\mathbb R$ through The canonical natural $\iota(n) = n \cdot 1_F$ of a field.