Prove the following series $\sum\limits_{s=0}^\infty \frac{1}{(sn)!}$ [duplicate]

It's $\sum\limits_{r=0}^{n-1} e^{i\frac{2\pi k}{n}r}=\frac{e^{i\frac{2\pi k}{n}n}-1}{e^{i\frac{2\pi k}{n}}-1}$ with $=0$ for $k\neq l\cdot n$ and $=n$ for $k=l\cdot n$, $l\in\mathbb{Z}$.

It follows $\sum\limits_{r=0}^{n-1} e^{e^{i\frac{2\pi}{n}r}}=\sum\limits_{k=0}^\infty\frac{1}{k!}\sum\limits_{r=0}^{n-1} e^{i\frac{2\pi k}{n}r}=n\sum\limits_{k=0}^\infty\frac{1}{(nk)!}$ and therefore the claim.

$\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

The roots of $\ds{z^{n} - 1 = 0}$ are given by $\ds{\braces{\exp\pars{{2\pi r \over n}\,\ic}\ \mid\ r = 0,1,\ldots,n - 1}}$. Note that $\ds{{1 \over n}\sum_{r = 0}^{n - 1}\exp\pars{{2\pi rs \over n}\,\ic}}$ is equal to $\ds{1}$ whenever $\ds{n \mid s}$ and it vanishes out otherwise.

\begin{align} \color{#f00}{\sum_{s = 0}^{\infty}{1 \over \pars{sn}!}} & = \sum_{s = 0}^{\infty}{1 \over s!}\, \bracks{{1 \over n}\sum_{r = 0}^{n - 1}\exp\pars{{2\pi rs \over n}\,\ic}} = {1 \over n}\sum_{r = 0}^{n - 1}\sum_{s = 0}^{\infty}{1 \over s!}\, \bracks{\exp\pars{{2\pi r \over n}\,\ic}}^{s} \\[5mm] & = {1 \over n}\sum_{r = 0}^{n - 1}\exp\pars{\exp\pars{{2\pi r \over n}\,\ic}} = {1 \over n}\sum_{r = 0}^{n - 1}\exp\pars{\cos\pars{2\pi r \over n}} \exp\pars{\ic\sin\pars{2\pi r \over n}} \\[5mm] & = {1 \over n}\sum_{r = 0}^{n - 1}\exp\pars{\cos\pars{2\pi r \over n}}\bracks{% \cos\pars{\sin\pars{2\pi r \over n}} + \sin\pars{\sin\pars{2\pi r \over n}}\ic}\label{1}\tag{1} \end{align}

It's clear that the imaginary part of \eqref{1} vanishes out.