<s>[INST] Enumerative Combinatorics problem from Stanley 
Let $n$ and $k$ be two positive integers. Show that
  \begin{align}
\sum a_1\cdot a_2 \cdot \cdots \cdot a_k={n+k-1\choose 2k-1} ,
\end{align}
  where the sum ranges over all compositions $(a_1,a_2,\ldots,a_k)$ of $n$ into $k$ parts.

Reminder: A composition of $n$ into $k$ parts is defined as a $k$-tuple $(a_1,a_2,\ldots,a_k)$ of positive integers such that $a_1 + a_2 + \cdots + a_k = n$.
This is Exercise 1.29 in Stanley's Enumerative Combinatorics volume 1 (2nd edition). I have no idea how to do it.
 [/INST] It suffices to prove they have the same generating function: 
\begin{align*}
\sum\limits_{n} \sum a_1\cdots a_k x^n &= \sum\limits_{1\leq a_1,\ldots,a_k} a_1\cdots a_k x^{a_1 + \ldots + a_k} \\
&= \sum\limits_{1\leq a_1,\ldots,a_k} (a_1 x^{a_1}) \cdots (a_k x^{a_k}) \\
&= \left(\sum\limits_{1\leq a_1} a_1 x^{a_1}\right)\cdots\left(\sum\limits_{1\leq a_k} a_k x^{a_k}\right) \\
&= \left( \frac{x}{(1 - x)^2} \right)^k \\
&= x^k \left( \frac{1}{(1 - x)^{2k}}\right) \\
&= x^k \sum\limits_m \binom{m + 2k - 1}{m}x^m \\
&= x^k \sum\limits_n \binom{n - k + 2k - 1}{n - k}x^{n-k} \\
&= \sum\limits_{n} \binom{n + k - 1}{2k - 1}x^n.
\end{align*}
</s>