Power series relation Draks gave the identity, Higher Order Trigonometric Function $$\sum_{k=0}^\infty \frac{(-1)^k x^{km}}{(km)!}=\frac{1}{m}\sum_{k=0}^{m-1} \exp( e^{i\frac{2k+1}{m}\pi}x )$$ How can this be proven?
 A: Writing $\zeta_{2m}$ for a $2m$th root of unity, $e^{\large i\pi \frac{2k+1}{m}n}=e^{\large2\pi i(\frac{k}{m}n+\frac{1}{2m}n)}=\zeta_m^{kn}\zeta_{2m}^n$. Thus
$$\frac{1}{m}\sum_{k=0}^{m-1}\exp\left(e^{i\pi \frac{2k+1}{m}}x\right)=\frac{1}{m}\sum_{k=0}^{m-1}\sum_{n=0}^\infty\frac{e^{\pi i\frac{2k+1}{m}n}x^n}{n!}$$
$$=\sum_{n=0}^\infty\frac{x^n}{n!}\frac{\zeta_{2m}^n}{m}\sum_{k=0}^{m-1}\zeta_{m}^{kn}=\sum_{n=0}^\infty\frac{x^{nm}}{(nm)!}(-1)^n$$
via $\zeta^{nm}_{2m}=(-1)^n$ and the orthogonality relation
$$\frac{1}{m}\sum_{k=0}^{m-1}\zeta_m^{kn}=\begin{cases}1 & m\mid n \\ 0 & m\nmid n.\end{cases} $$
A: Let $x_k=e^{i\frac{2k+1}{2m}2\pi}$. These are the $2m$th roots of unity that are not $m$th roots of unit, so they are the $m$ distinct roots of roots of $x^m+1=0$.
Now substituting above, and expanding $\exp$ we get:
$$\frac{1}{m} \sum_{k=0}^{m-1} \exp(x_k x) = \sum_{k=0}^{m-1} \sum_{j=0}^\infty \frac{x_k^jx^j}{j!} = \sum_{j=0}^{\infty} \frac{x^j}{j!}\sum_{k=0}^{m-1}x_k^j$$
So now you need to show that $\sum_{k=0}^{m-1} x_k^j=0$ if $m\not\mid j$, $-m$ if $\frac{j}{m}$ is odd, and $m$ if $\frac{j}{m}$ is even.
