Finite differences and alternating binomial sums

Step 1: $\Delta$ lowers the degree by $1$. If $P(x)=ax^d+\cdots$ with $d\ge1$, then $(x+1)^d-x^d=dx^{d-1}+\cdots$, so $\Delta P$ has degree $d-1$ and leading coefficient $da$. If $P$ is constant, $\Delta P=0$. By induction, if $\deg P=d$ then $\Delta^dP=a\,d!$ (a constant) and $\Delta^{m}P=0$ for $m>d$.

Step 2: a formula for $\Delta^n$. Let $E$ be the shift $(EP)(x)=P(x+1)$ and $I$ the identity. Then $\Delta=E-I$, and $E$, $I$ commute, so the binomial theorem applies:
$$\Delta^n=(E-I)^n=\sum_{k=0}^n(-1)^{n-k}\binom nkE^k,\qquad\text{so}\qquad(\Delta^nP)(0)=\sum_{k=0}^n(-1)^{n-k}\binom nkP(k).$$

Step 3. Multiplying by $(-1)^n$:
$$\sum_{k=0}^n(-1)^k\binom nkP(k)=(-1)^n(\Delta^nP)(0).$$
By Step 1 this is $0$ if $d<n$ and $(-1)^n\,n!\,a$ if $d=n$. $\blacksquare$

Check. For $P(x)=x^m$ with $n=4$: the sums for $m=0,\dots,4$ are $0,0,0,0,24$.

Lesson. $\Delta$ is "differentiation for sequences": it lowers the degree by one, and $n$ applications expand into an alternating binomial sum.