I've been manipulating the series definitions of the Taylor (Maclaurin) and higher-order derivatives.
The definitions I'm working with are as follows:
\begin{align*} T(x) &\equiv \sum_{n=0}^\infty\frac{D^nf(0)}{n!}x^n,\\ D^nf(x)&\equiv \lim_{h\to0}\sum _{k=0}^n \binom nk \frac{(-1)^k}{h^n}f(x+(n-k)h).\end{align*}
Meaning that \begin{align*}f(x) &= \lim_{h \to 0} \sum_{n=0}^\infty\frac{x^n}{n!}\sum _{k=0}^n\binom nk \frac{(-1)^k}{h^n}f((n-k)h)\\ &= \lim_{h \to 0} \sum_{n=0}^\infty\sum _{k=0}^n\frac{(-1)^k x^n}{k!(n-k)!h^n}f((n-k)h)\\ &= \lim_{h \to 0} \sum_{n=0}^\infty\sum _{k=0}^n\frac{(-1)^k x^k}{k!h^k} \frac{x^{n-k}f(h(n-k))}{(n-k)!h^{n-k}},\end{align*} from which I reasoned that with Cauchy's formula, $ \sum a_n \sum b_n = \sum_{n=0}^\infty\sum_{k=0}^na_kb_{n-k} $: \begin{align*}f(x) &= \lim_{h \to 0} \sum_{k=0}^\infty\frac{(-1)^k x^k}{k!h^k}\sum _{n=0}^\infty\frac{x^nf(hn)}{n!h^{n}}\\&= \lim_{h \to 0} e^{-x/h}\sum _{n=0}^\infty\frac{x^nf(hn)}{n!h^{n}}.\end{align*}
