I want to rewrite the following product of sums in $x$
$$Y=\sum_{n=0}^\infty\sum_{m=0}^\infty\sum_{k=0}^\infty \frac{i^{n+m+k}(-1)^{n+m}}{(n+1)!(m+1)!k!} x^{n+m+k}$$
as a single sum in $x^s$. It seems to me like the problem is essentially, for a given $s=n+m+k$, to list all values of (n,m,k) that sum to $s$ and sum the corresponding coefficients of the expression above. So $$Y=\sum_s^\infty i^s a_s x^s,$$ where $$a_s = \left.\sum_{\{m,n,k\}}\right|_{m+n+k=s} \frac{(-1)^{n+m}}{(n+1)!(m+1)!k!}.$$ I'm not really sure how to go about this though.. I know for sure that $a_s$ can be written in compact form. This is not homework, I am trying to reproduce some results from a paper.
$$Y=\sum_{n=0}^\infty\sum_{m=0}^\infty\sum_{k=0}^\infty \dfrac{i^{n+m+k}(-1)^{n+m}}{(n+1)!(m+1)!k!} x^{n+m+k}\\ =\left(\dfrac{1}{ix}\sum_{n=0}^\infty\dfrac{(-ix)^{n+1}}{(n+1)!}\right)\left(\dfrac{1}{ix}\sum_{m=0}^\infty\dfrac{(-ix)^{m+1}}{(m+1)!}\right)\left(\sum_{k=0}^\infty\dfrac{(ix)^k}{k!}\right)\\ =\boxed{\dfrac{(e^{-ix}-1)^2e^{ix}}{i^2x^2}}$$