Basically I want to see if I can use induction in the following way:
Statement:
$$\sum_{p=1}^{n}\binom{n}{p}p^m(-1)^p=0, m \in \mathbb{N}, 1\leq m < n $$
Proof:
Let m=1, $n>1$
\begin{align*} \sum_{p=1}^{n}\binom{n}{p}p^1(-1)^p &= \sum_{p=1}^{n}\frac{n}{p}\binom{n-1}{p-1}(-1)^pp \\ &= n\sum_{p=1}^{n}\binom{n-1}{p-1}(-1)^p \\ &= n\sum_{p=0}^{n-1}\binom{n-1}{p}(-1)^{p+1} \\ &= -n\sum_{p=0}^{n-1}\binom{n-1}{p}(-1)^p \\ &= -n\sum_{p=0}^{n-1}\binom{n-1}{p}1^{n-1-p}(-1)^p \\ &= -n(1-1)^{n-1} \\ &= -n(0)^{n-1} \\ &= 0 \end{align*}
So the base case holds.
Now, let m=k, $n>k$ and assume the following:
$$\sum_{p=1}^{n}\binom{n}{p}p^k(-1)^p=0$$
Now, let $m=k+1$, $n>k+1$:
\begin{align*} \sum_{p=1}^{n}\binom{n}{p}p^{k+1}(-1)^p &= \sum_{p=1}^{n}\frac{n}{p}\binom{n-1}{p-1}p^{k+1}(-1)^p \\ &= n\sum_{p=1}^{n}\binom{n-1}{p-1}p^k(-1)^p \\ &= n\sum_{p=0}^{n-1}\binom{n-1}{p}(p+1)^k(-1)^{p+1} \\ &= -n\sum_{p=0}^{n-1}(-1)^p(p^k+kp^{k-1}+....+kp+1)\binom{n-1}{p} \\ &= -n\sum_{p=0}^{n-1}\binom{n-1}{p}(-1)^pp^k+\binom{n-1}{p}(-1)^pkp^{k-1} \\ & +...+\binom{n-1}{p}(-1)^pkp+\binom{n-1}{p}(-1)^p \\ &= -n[\{\sum_{p=0}^{n-1}\binom{n-1}{p}(-1)^pp^k\}+\{\sum_{p=0}^{n-1}\binom{n-1}{p}(-1)^p* \\ & kp^{k-1}\}+...+\{\sum_{p=0}^{n-1}\binom{n-1}{p}(-1)^pkp\}+\{\sum_{p=0}^{n-1}\binom{n-1}{p}(-1)^p \\ & \}] \\ &= -n[\{\sum_{p=1}^{n-1}\binom{n-1}{p}(-1)^pp^k\}+\{\sum_{p=1}{n-1}\binom{n-1}{p}(-1)^p* \\ & kp^{k-1}\}+...+\{\sum_{p=1}^{n-1}\binom{n-1}{p}(-1)^pkp\}+\{\sum_{p=0}^{n-1}\binom{n-1}{p}* \\ & (-1)^p\}] \end{align*} Now, we assumed that $n>k+1$, therefore $n-1>k$. Hence, $n-1>k-1$, hence $n-1>k-a$ for all natural numbers $a$. Earlier I assumed that the equality holds if the upper bound (n, in that case) is greater than the exponent of p in that sum (in that case, k). Here, our upper bound is n-1 and our exponent is $k-a$. So if I can use unductive logic here: $$\sum_{p=1}^{n}\binom{n}{p}p^{k+1}(-1)^p=-n[0+0+0+0+......+0]=0$$ That should complete the proof. Is my logic correct?
