TITLE: If $P(x)$ is any polynomial of degree less than $n$, show that $\sum_{j=0}^n (-1)^j\binom{n}{j}P(j)=0$.
QUESTION [6 upvotes]: If $P(x)$ is any polynomial of degree less than $n$, then prove that
$$\sum_{j=0}^n (-1)^j\binom{n}{j}P(j)=0$$

My approach was to try and prove this separately for $j^k\ \ \forall\ \ k<n$, instead of $P(j)$, since $P(x)$ can be written as $a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+...+a_0$. However, I could not get anywhere.It would be much appreciated if someone could provide a rigorous solution for this or any key insights.Thanks a lot

REPLY [2 votes]: For a combinatorial proof of the sum with $j^k$, note that both sides count the number of surjections from an $k$-set onto an $n$-set, which is $0$ when $k<n$. The summation is the inclusion-exclusion formula, where the $n$ properties to be avoided are that element $i$ is not in the image of the function.