Using the Multinomial Theorem to Calculate a Finite Sum raised to an exponent I know it's a simple question, but I keep getting different general formulas for the coefficients when I am trying to use the multinomial theorem for the following:
$$
\left(\sum_{k=0}^{M}\frac{(-x^2)^k}{4^kn^{k/2}k!(1+k)!}\right)^n
$$
Please help me to calculate it.
Thank you.
 A: $\newcommand{\+}{^{\dagger}}%
 \newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
 \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}%
 \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}%
 \newcommand{\dd}{{\rm d}}%
 \newcommand{\isdiv}{\,\left.\right\vert\,}%
 \newcommand{\ds}[1]{\displaystyle{#1}}%
 \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}%
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}%
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}%
 \newcommand{\ic}{{\rm i}}%
 \newcommand{\imp}{\Longrightarrow}%
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}%
 \newcommand{\pars}[1]{\left( #1 \right)}%
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}%
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}%
 \newcommand{\sech}{\,{\rm sech}}%
 \newcommand{\sgn}{\,{\rm sgn}}%
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}%
 \newcommand{\verts}[1]{\left\vert #1 \right\vert}%
 \newcommand{\yy}{\Longleftrightarrow}$
$\ds{%
\bracks{\,\sum_{k = 0}^{M}{(-x^{2})^{k} \over 4^{k}n^{k/2}k!(1 + k)!}}^{n}:\
{\large ?}.\quad}$ Let $\ds{\quad z \equiv {-x^{2} \over 4n^{1/2}}\quad}$ and
$\ds{\quad a_{k} \equiv {1 \over k!\pars{1 + k}!}}$.

$$
\mbox{Then}\,,\quad
\bracks{\,\sum_{k = 0}^{M}{(-x^{2})^{k} \over 4^{k}n^{k/2}k!(1 + k)!}}^{n}
=
\pars{\sum_{k = 0}^{M}a_{k}z^{k}}^{n}\tag{1}
$$

\begin{align}
&\pars{\sum_{k = 0}^{M}a_{k}z^{k}}^{n}=
\sum_{k_{1} = 0}^{M}a_{k_{1}}z^{k_{1}}\sum_{k_{2} = 0}^{M}a_{k_{2}}z^{k_{2}}
\cdots\sum_{k_{n} = 0}^{M}a_{k_{n}}z^{k_{n}}\sum_{\ell = 0}^{nM}
\delta_{\ell,k_{1}\ +\ k_{2}\ +\ \cdots\ +\ k_{n}}
=
\sum_{\ell = 0}^{nM}A_{\ell}z^{\ell}
\end{align}
where

$$
\left.A_{\ell}
\equiv
\sum_{k_{1}, k_{2},\ldots,k_{n} = 0}^{M}
a_{k_{1}}a_{k_{2}}\ldots a_{k_{n}}\right\vert_{\sum\limits_{i\ =\ 1}^{n}k_{i}\ =\ \ell}
$$

