How does this manipulation of summations work? I am reading some mathematics in which is the following algebraic manipulation.
$$
\begin{align}
\exp(x)\exp(y) & = \left(\sum_{n = 0}^\infty \frac{x^n}{n!}\right) \left(\sum_{m = 0}^\infty \frac{y^m}{m!}\right) \\
& = \sum_{n, m = 0}^\infty \frac{x^n y^m}{n! m!} \\
& = \sum_{n, m = 0}^\infty \frac{(n + m)!}{n! m!} x^n y^m \frac{1}{(n + m)!} \\
& = \sum_{k = 0}^\infty \sum_{n = 0}^k \left[\frac{k!}{n!(k - n)!}x^n y^{k - n} \right] \frac{1}{k!} \\
\end{align}
$$
I haven't yet understood the step from the third line to the fourth line.
In particular, where does the $\sum_{n = 0}^k$ in the fourth line appear from?
I would appreciate help to understand this.
NB: I assume that $\sum_{n, m = 0}^\infty \frac{x^n y^m}{n! m!}$ is shorthand for $\sum_{n = 0}^\infty \sum_{m = 0}^\infty \frac{x^n y^m}{n! m!}.$
I think that this assumption is correct, but please let me know if I am wrong.
 A: Since the sum is absolutely convergent, we can sum in any order.
In the fourth line, we just sum along the diagonals $n+m = \text{constant}$.
This is known as a Cauchy product: http://en.wikipedia.org/wiki/Cauchy_product
A: $$
\sum_{n,m=0}^\infty A(n,m,n+m) = \sum_{k=0}^\infty\sum_{n=0}^k A(n,n-k,k)
$$
$$
\begin{array}{c|ccccccccc}
& m=0 & m=1 & m=2 & m=3 & \cdots \\
\hline
n=0 & \bullet & \bullet & \bullet & \circ & \cdots \\
n=1 & \bullet & \bullet & \circ & \cdots \\
n=2 & \bullet & \circ & \cdots \\
n=3 & \circ & \cdots \\
\vdots & \vdots
\end{array}
$$
Look at the diagonal where "$\circ$" appears instead of "$\bullet$".  Those are the entries in which $k=3$.  As you move downward and to the left along that diagonal, you have $n=0$, then $n=1$, then $n=2$, then $n=3$; hence $\displaystyle\sum_{n=0}^3=\sum_{n=0}^k$.
Notice that in every entry on that diagonal you have $n+m=3$, i.e. $n+m=k$.  So in the expression $A(n,m,n+m)$, where $n+m$ appears, it is replaced by $k$, in $A(n,n-k,k)$.  Since $n+m=k$, we have $m=k-n$, so where $m$ appears in places where it is not in the expression $n+m$, we put $k-n$ in its place.
