Consider counting $f_n(l)$, the words of length $n$ with exactly $l$ letters.
I tried to find a recursive formula. I came up with
$$ f_n(l) = l^n - \sum_{j=0}^{l-1} \binom l j f_n(j) $$
Since we count $l^n$ words with no restrictions on which letters are used, then subtract away words with exactly $j$ letters, for $ j < l$.
I worked out a few cases for small $l$ and I found out the coefficients were binomial coefficients with alternating signs (I found out it can be derived using inclusion-exclusion). That is
$$ f_n(l) = \sum_{k=0}^l (-1)^{l-k} \binom l k k^n $$
If we look at the coefficient of $k^n$ in each expression, we get for $k < l$:
$$ -\sum_{j=0}^{l-1} \binom l j (-1)^{j-k} \binom j k = (-1)^{l-k} \binom l k $$
Is there an easy way to prove this expression? Can this technique be used to prove any other identities?
