combinatorics: sum of product of integer compositions I am trying to solve a problem from Stanley's book, it says:
Fix $k,n \in \mathbb{P}$. Show that:
\begin{align}
\sum a_1 a_2 \cdots a_k = \binom{n+k-1}{2k-1}
\end{align}
where the sum ranges over all compositions $(a_1 , a_2 , \ldots , a_k)$ of $n$ into $k$ parts.
I am trying to reason like this:
we need to find the coefficient $c_n = \sum a_1 a_2 \cdots a_k$ from this generating function --
\begin{align}
\sum_n c_n x^n &= \sum_n \sum a_1 a_2 \cdots a_k x^n \\ 
&= \sum_n \sum a_1 a_2 \cdots a_k x^{a_1 + a_2 + \cdots + a_k}\\
&= \sum_n \sum a_1x^{a_1} a_2x^{a_2} \cdots a_kx^{a_k}
\end{align}
after that, I have no clue, how do I solve this ? 
moreover, what is the range in the inner sum ?

If we consider Mark Riedel's answer, and assume $n=4$, $k=2$; then the sum will be 
\begin{align}
\sum (z + 2z^2)^2 = z^2 + 4z^3 + 4z^4
\end{align}
On the other hand the compositions will be $(1,3), (2,2), (3,1)$, therefore the above sum will be counted as:
\begin{align}
(1.3)z^{1+3} + (2.2)z^{2+2} + (3.1)z^{3+1} &= 1z^1.3z^3 + 2z^2.2z^2 + 3z^3.1z^1\\ 
&= 3z^4 + 4z^4 + 3z^4 = 10z^4
\end{align} 
what's going on? what am I missing?

We have using generating functions the closed form
$$[z^n](z+2z^2+3z^3+\cdots)^k
= [z^n] \frac{z^k}{(1-z)^{2k}}
\\ = [z^{n-k}] \frac{1}{(1-z)^{2k}}.$$
Using the Newton binomial this becomes
$${n-k+2k-1\choose 2k-1} = {n+k-1\choose 2k-1}.$$
