Show that $\sum_{k=0}^{n} \binom{n}{k} ka^k = an(a+1)^{n-1}$ Problem:
Show that $\sum_{k=0}^{N} \binom{N}{k} ka^k = aN(a+1)^{N-1}$
Attempt:
I tried using induction but got stuck.
$N=0$ implies $\sum_{k=0}^{0} \binom{0}{k} ka^k = 0$
$N=1$ implies $\sum_{k=0}^{1} \binom{1}{k} ka^k = a = a(1)(a+1)^{1-1}$
$N=2$ implies $\sum_{k=0}^{2} \binom{2}{k} ka^k = 2a(a+1)^{2-1}$
Assume that for $N=n$, $\sum_{k=0}^{n} \binom{n}{k} ka^k = na(a+1)^{n-1}$. Then, for $N=n+1$, we find
$$\sum_{k=0}^{n+1} \binom{n+1}{k} ka^k = \sum_{k=0}^{n+1} \left(\binom{n}{k-1}+\binom{n}{k}\right) ka^k$$
Splitting up the sum on the right gives
$$\sum_{k=0}^{n+1} \binom{n}{k-1}ka^k + \sum_{k=0}^{n} \binom{n}{k}ka^k + \binom{n}{n+1}ka^k$$
and since $\binom{n}{n+1}=0$, we have
$$\sum_{k=0}^{n+1} \binom{n+1}{k} ka^k = \sum_{k=0}^{n+1} \binom{n}{k-1}ka^k + an(a+1)^{n-1}$$
and this is where I'm stuck. If we look at it from the other direction, we have
$$(n+1)a(a+1)^{n+1-1} = (n+1)a(a+1)^n = na(a+1)^n + a(a+1)$$
But I'm not sure where to go from here.

$$S=\sum_{k=0}^{n} k {n \choose k} a^k =\sum_{k=0}^{n} \frac{n!}{k! (n-k)!} a^k= \sum_{k=0}^{n} \frac{n (n-1)!}{(k-1)! (n-k)!}=n\sum_{k=0}^{n} {n-1 \choose k-1} a^k$$
Let $k-1=p$, then
$$S=n\sum_{p=-1}^{n-1} {n-1 \choose p} a^{p+1}=na\sum_{p=0}^{n-1} {n-1 \choose p} a^p=na(1+a)^{n-1}.$$
Note that ${m \choose -n}=0$, where $n,m$ are positive integers.
