Suppose that $$H_{N,k}(x)=\frac{x^ke^{\frac{-x}{N}}}{N^{k-1}k!\sum_{n=0}^{N-1}{w_N^{-nk}e^{\frac{w_N^nx}{N}}}}=\sum_{n=0}^\infty{A_n\frac{x^n}{n!}}$$
where $k\lt N, w_N=e^{\frac{2i\pi}{N}}$, and both are natural numbers. I would like to show that, if $N$ is even, then
$$A_{k}=\frac{1}{N^kk!}; A_{Nj+k}=0, j>0$$
Because of the nature of the coefficients, since I want to prove the zero terms are periodic, I was using the multisection theorem that states, if a function is analytic, then the multisection of a power series of a function has the closed form
$$M_{N,k}(x)=\sum_{j=-\infty}^\infty{A_{Nj+k}x^{Nj+k}}=\frac{1}{N}\sum_{j=0}^{N-1}w_N^{-jk}H_{N,k}(w_N^jx)$$
Thus, to prove the above, I would just use the multisection closed form and show that it equals $\frac{1}{N^kk!}$, as I have other problems that show that if $Nj+k<N$, then $A_k$ is this particular value, thus implying that the remaining coefficients are $0$. So
$$M_{N,k}(x)=\frac{1}{N}\sum_{j=0}^{N-1}w_N^{-jk}H_{N,k}(w_N^jx)$$ $$=\frac{1}{N}\sum_{j=0}^{N-1}w_N^{-jk}\frac{(w_N^jx)^ke^{\frac{-w_N^jx}{N}}}{N^{k-1}k!\sum_{n=0}^{N-1}{w_N^{-nk}e^{\frac{w_N^nw_N^jx}{N}}}}$$ $$=\frac{x^k}{N^kk!}\sum_{j=0}^{N-1}\frac{e^{\frac{-w_N^jx}{N}}}{\sum_{n=0}^{N-1}{w_N^{-nk}e^{\frac{w_N^{n+j}x}{N}}}}$$ So the idea then is this; since $N$ is even, the above exponent of the exponential is just another root of unity, which would not be the case for odd $N$, so if $N=2m$, then $-w_N^j=w_N^{j+m}$, I can see that I'm closing in of the fact that the sum is getting closer to becoming $1$, but can anyone help me see past the next step I'm missing? I've been staring at it and just can't get past it. Perhaps factoring a particular root of unity from the denominator and then using its inverse in the numerator. I know i'm close ust and just struggling with the last couple of steps.