How to simplify $f(x)=\sum\limits_{i=0}^{\infty}\frac{x^{i \;\bmod (k-1)}}{i!}$? $$f(x)=\sum_{i=0}^{\infty}\frac{x^{i \;\bmod (k-1)}}{i!}$$
${i \bmod (k-1)}$ $\quad$ says the $x$ powers can be only $x^0$, $x^1$, ...,$x^{k-2}$
Understand simplify a way to transform this infinity sum in a finite formula.
 A: Group them by $0,1,2$ and so on. Your formula is 
$$\sum_{j=0}^{k-2} \left( x^j \sum_{n=0}^\infty \frac{1}{(n(k-1)+j)!}  \right) \,.$$
You should be able to get a closed formula for inside bracket by calculating the Taylor series of $e^z$ at the $(k-1)$-roots of unity. 
P.S. For the last part:  adding $e^{w_1}+...+e^{w_{k-1}}$ yields the coefficient of $x^0$. Now if you add $\sum w_ie^{w_i}$ you get the coefficient of $x^1$, and so on.
A: This is an expansion of the other answer. There's no point in working modulo $k-1$ instead of $k$.
$$\sum_{m=0}^\infty\frac{x^{(m\mod k)}}{m!}=\sum_{m=0}^{k-1}\left(\frac{1}{m!}+\frac{1}{(m+k)!}+\frac{1}{(m+2k)!}+\cdots\right) x^m.$$
Now if $\omega$ is a primitive $k$-th root of unity we have the property
$$\frac{1}{k}\sum_{l=0}^{k-1}\omega^{jl}=\begin{cases}1&j\equiv0\mod k;\\0&j\not\equiv0\mod k.\end{cases}$$
Thus
$$\frac{1}{k}\sum_{l=0}^{k-1}\exp(\omega^l)\omega^{-ml}=\frac{1}{k}\sum_{l=0}^{k-1}\left(\sum_{n=0}^\infty\frac{(\omega^l)^n}{n!}\right)\omega^{-ml}=\sum_{n=0}^\infty\frac{1}{n!}\left(\frac{1}{k}\sum_{l=0}^{k-1}\omega^{(n-m)l}\right).$$
The inner sum above is $1$ when $n\equiv m\mod k$ and $0$ otherwise so this equals
$$\frac{1}{m!}+\frac{1}{(m+k)!}+\frac{1}{(m+2k)!}+\cdots.$$
Therefore our original sum is
$$(\bullet)=\frac{1}{k}\sum_{m=0}^{k-1}\sum_{l=0}^{k-1}\exp(\omega^l)(x\omega^{-l})^m.$$
