Taking a power of a power series, inside of a power series I have a power series nestled inside a power series, and the inside power series is taken to a power. I need to isolate the term that would go like $t^{2mn}$ so that I could integrate it.
I'm not looking to sum up the series, since that's where I'm coming from.
$$ \sum_{n=0}^{\infty} (-1)^n \left(\sum_{m=0}^{\infty} \; \frac{(-1)^m}{(2m)!} \; \beta^{2m} t^{2m} \right)^n $$
Any ideas are appreciated.
 A: Here are  two   approaches to extract the coefficient of $t^{2MN}$. One is based upon multinomial expansion, the other is based upon the representation $\cos t=\frac{1}{2}\left(e^{it}+e^{-it}\right)$. Regrettably, none of them seems really promising. Nevertheless, here we go!
It is convenient to use the coefficient of operator $[t^n]$ to denote the coefficient of $t^n$.

First approach:
We obtain
  \begin{align*}
\color{blue}{[t^{2MN}]}&\color{blue}{\sum_{n=0}^\infty(-1)^n\left(\sum_{m=0}^\infty\frac{(-1)^m}{(2m)!}\beta^{2m}t^{2m}\right)^n}\\
&\sum_{n=0}^\infty(-1)^n[t^{2MN}]\left(\sum_{m=0}^{MN}\frac{(-1)^m}{(2m)!}\beta^{2m}t^{2m}\right)^n\\
&=\sum_{n=0}^\infty(-1)^n[t^{2MN}]\sum_{{k_0+\cdots+k_{MN}=n}\atop{k_j\geq 0,\ ,0\leq j\leq MN}}
\binom{n}{k_0,\ldots,k_{MN}}\prod_{j=0}^{MN}\left(\frac{\left(-\beta^2\right)^j}{(2j)!}t^{2j}\right)^{k_j}\\
&\,\,\color{blue}{=\sum_{n=0}^\infty(-1)^n[t^{2MN}]\sum_{{k_0+\cdots+k_{MN}=n}\atop{k_j\geq 0,\ ,0\leq j\leq MN}}\binom{n}{k_0,\ldots,k_{MN}}}\\
&\qquad\quad\color{blue}{\cdot\prod_{j=0}^{MN}\left((2j)!\right)^{-k_j}\left(-\beta^2\right)^{\sum_{j=1}^{N}jk_j}t^{2\sum_{j=1}^{N}jk_j}}\tag{1}\\
\end{align*}
We  observe in  order   to find  the  coefficient   of   $t^{2MN}$  the expression (1) boils down to find the solutions  of
  \begin{align*}
k_0+k_1+\cdots+k_{MN}&=n\\
k_1+2k_2+\cdots+MNk_{MN}&=MN\qquad\qquad k_j\geq 0,  \,\, 0\leq j\leq  MN
\end{align*}
  which   indicates    connections  with  Bell  polynomials.

The second approach is somewhat more promising, since we can completely extract the coefficient of $t^{2MN}$ . But, we will see the final expression is far from being simple.

Second approach:
\begin{align*}
\color{blue}{[t^{2MN}]}&\color{blue}{\sum_{n=0}^\infty(-1)^n\left(\sum_{m=0}^\infty\frac{(-1)^m}{(2m)!}\beta^{2m}t^{2m}\right)^n}\\
&=[t^{2MN}]\sum_{n=0}^\infty(-1)^n\cos^n(\beta t)\tag{2}\\
&=[t^{2MN}]\sum_{n=0}^\infty(-1)^n\frac{1}{2^n}\left(e^{i\beta t}+e^{-i\beta  t}\right)^n\tag{3}\\
&=[t^{2MN}]\sum_{n=0}^\infty\left(-\frac{1}{2}\right)^ne^{-in\beta t}\left(1+e^{2i\beta t}\right)^n\tag{4}\\
&=[t^{2MN}]\sum_{n=0}^\infty\left(-\frac{1}{2}\right)^n\sum_{j=0}^\infty\frac{\left(-in\beta\right)^j}{j!}t^j\left(1+e^{2i\beta t}\right)^n\tag{5}\\
&=\sum_{n=0}^\infty\left(-\frac{1}{2}\right)^n\sum_{j=0}^{2MN}\frac{\left(-in\beta\right)^j}{j!}[t^{2MN-j}]\left(1+e^{2i\beta t}\right)^n\tag{6}\\
&=\sum_{n=0}^\infty\left(-\frac{1}{2}\right)^n\sum_{j=0}^{2MN}\frac{\left(-in\beta\right)^{2MN-j}}{(2MN-j)!}[t^{j}]\sum_{k=0}^n\binom{n}{k}e^{2ik\beta t}\tag{7}\\
&=\sum_{n=0}^\infty\left(-\frac{1}{2}\right)^n\sum_{j=0}^{2MN}\frac{\left(-in\beta\right)^{2MN-j}}{(2MN-j)!}
[t^{j}]\sum_{k=0}^n\binom{n}{k}\sum_{l=0}^\infty \frac{\left(2ik\beta\right)^l}{l!}t^l\tag{8}\\
&=\sum_{n=0}^\infty\left(-\frac{1}{2}\right)^n\sum_{j=0}^{2MN}\frac{\left(-in\beta\right)^{2MN-j}}{(2MN-j)!}\sum_{k=0}^n\binom{n}{k}\frac{(2ik\beta)^j}{j!}\tag{9}\\
&\,\,\color{blue}{=\frac{\left(-\beta^2\right)^{MN}}{(MN)!}\sum_{n=0}^\infty\left(-\frac{1}{2}\right)^nn^{2MN}
\sum_{j=0}^{2MN}\binom{2MN}{j}\left(-\frac{2}{n}\right)^j\sum_{k=0}^n\binom{n}{k}k^j}\tag{10}
\end{align*}

Comment:


*

*In (2) we use the series expansion $\cos t=\sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}t^{2n}$.

*In (3) we use the representation $\cos t=\frac{1}{2i}\left(e^{it}+e^{-it}\right)$.

*In (4) we factor out $e^{-in\beta t}$.

*In (5) we expand $e^{-in\beta t}$.

*In (6) we apply the rule $[t^{p-q}]A(t)=[t^p]t^qA(t)$ and we  set the  upper  limit of the sum with   index  $j$ to  $2MN$, since other terms do not contribute.

*In (7) we change the order of the  inner  sum $j\to   2MN-j$.

*In  (8) we  expand $e^{2ik\beta t}$.

*In (9) select the coefficient of $t^j$.

*In (10) we  do  some  final rearrangements.
A: Seems like a strange approach but the only thing I think would help would be this: https://en.m.wikipedia.org/wiki/Cauchy_product
Although you'd need to establish a general form for the coefficients when multiplying more than 2 series, which would include several nested finite sums, making the result really ugly
