I recently encountered the following
claim. let $d$ be a positive integer and $f(t)=a_0+\dots+a_{d-1}t^{d-1}$ be a polynomial of degree less than $d$. then $\displaystyle\sum_{j=0}^d (-1)^d {d\choose j}f(j)=0$.
and proved it by considering a Vandermonde matrix, with an added row of the values of $f$.
($a_{i,j}=j^i$, for $i=0,\dots,d-1$ and $a_{d,j}=f(j)$, $j=0,\dots,d$).
Question. is there a simpler (without determinants) proof for this identity?
I think this is the most elementary proof: for any polynomial $f$, define its discrete derivative $\Delta f(x) = f(x+1) - f(x)$. It is easy to see that the degree of $\Delta f$ is precisely one less than that of $f$. Iterating this observation, we see that $\Delta^{d+1} f$ is the zero function. One can prove inductively that its values are precisely given by the sum you are trying to prove is identically zero, but I will leave the details to you.
Consider the polynomials $$P_k(t)=\prod_{\substack{j=0\\ j\neq k}}^d (t-j), \quad k=0,1,\ldots, d $$ Then $\deg P_k=d.$ These polynomials form a linear basis for the space of polynomials of degree less than or equal to $d,$ because the matrix $\{P_k(l)\}_{k,l=0}^d$ is diagonal with nonzero entries on the main diagonal.
The family of polynomials $$Q_k(t)=P_k(t)-P_d(t)\quad k=0,1,\ldots, d-1$$ is thus a basis for the space of polynomials of degree less than or equal to $d-1.$
Observe that $$P_k(j)=\begin{cases} (-1)^{d-k}\,k!\,(d-k)! & j=k\\ 0 & j\neq k \end{cases}$$ Hence $$\sum_{j=0}^d(-1)^j {d\choose j} P_k(j)=(-1)^k{d\choose k}(-1)^{d-k}\,k!\,(d-k)!=(-1)^d d! $$ Therefore $$\sum_{j=0}^d(-1)^j {d\choose j} Q_k(j)=0$$ As the formula holds for a basis, then it extends to the entire space of polynomials of degree less than or equal to $d-1.$
The method described above gives rise to generalizations. Let $t_0<t_1<\ldots <t_d.$ Then for any polynomial $P(t),$ such that $\deg P\le d-1,$ we have $$\sum_{j=0}^d\left [ \prod_{\substack{k=0\\ k\neq j}}^d (t_k-t_j)^{-1}\right ]\, P(t_j)=0 \qquad (*)$$ For $t_j=j$ we get $$\prod_{\substack{k=0\\ k\neq j}}^d (t_k-t_j)^{-1}={(-1)^j\over j!\, (d-j)!}$$ The proof of $(*)$ can be also performed by suitably defined discrete derivatives similar to the ones in Answer $1$.
proof. by linearity we may assume $f(t)={t\choose k}$ for some $k<d$. then this reduces to the identity $\sum_{j}(-1)^{j}{d \choose j}{j \choose k}=0$ which holds by the following combinatorial argument. first we choose $k$ out of $d$ elements, and then we choose a subset of the remaining elements, only the sign in our count depends on the parity of the size of the subset.
– tomm Feb 26 '22 at 23:18