convolutions of formal series I'm a little confused on how convolution of formal series works mostly the indexing. This is the problem I'm working on. 
Let $n$ and $k$ be fixed. 
Calculate
$$\sum_{i=0}^{n}(-1)^{i}\binom{n}{i}i^{k}$$
My idea was to use generating functions I got 
$$A(x)=\sum_{n\geq 0}\left(\sum_{i=0}^{n}(-1)^{i}\binom{n}{i}i^{k}\right)x^{n}$$
This looks like the convolution of two series, $(-1)^{i}i^{k}$ and $\binom{n}{i}$. 
But I'm confused on the indexing. Would
$A(x)=B(x)C(x)$ where $B(x)=\sum_{n\geq0}\binom{n}{n-i}x^{n}$ and $C(x)=\sum_{n\geq0}(-1)^{n}n^{k}$? I think $B(x)$ is incorrect but what should be the index for $B(x)$? 
Thanks. 
 A: Permit me to re-write this as
$$(-1)^n \sum_{k=0}^n {n\choose k} (-1)^{n-k} k^m$$
because I use $i$ in complex variable calculations.
Then we see immediately that
$$A(z) = \sum_{n\ge 0} (-1)^n \frac{z^n}{n!} = \exp(-z).$$
Furthermore
$$B(z) = \sum_{n\ge 0} n^m \frac{z^n}{n!} = 
z \exp(z) \sum_{q=1}^m {m \brace q} z^{q-1},$$
which may be verified by induction. Here we have used the Stirling numbers of the second kind.
It follows that $$A(z) B(z) = z \sum_{q=1}^m {m \brace q} z^{q-1}.$$
Therefore the sum is
$$(-1)^n n! [z^n] A(z) B(z) = 
(-1)^n n! [z^{n-1}] \sum_{q=1}^m {m \brace q} z^{q-1}
= (-1)^n n! {m \brace n}.$$
Addendum. The induction proof of the formula for $B(z)$ goes like this. Suppose
$$B_m(z) = \exp(z) \sum_{q=1}^m {m \brace q} z^q.$$
Then $$B_{m+1}(z) = z \frac{d}{dz} B_m(z),$$ which gives
 $$B_{m+1}(z) = \exp(z) \sum_{q=1}^m {m \brace q} z^{q+1}
+ \exp(z) \sum_{q=1}^m {m \brace q} q z^q\\
= \exp(z)\left({m \brace m} z^{m+1} +
\sum_{q=1}^m \left(q {m \brace q} + {m \brace q-1}\right)z^q \right)
\\= \exp(z) \sum_{q=1}^{m+1} {m+1 \brace q} z^q$$
by the defining recurrence of the Stirling numbers of the second kind.
There is another calculation of this type at this MSE link -- I and at this MSE link -- II.
