A Curious Binomial Sum Identity without Calculus of Finite Differences Let $f$ be a polynomial of degree $m$ in $t$. The following curious identity holds for $n \geq m$,
\begin{align}
\binom{t}{n+1} \sum_{j = 0}^{n} (-1)^{j} \binom{n}{j} \frac{f(j)}{t - j} = (-1)^{n} \frac{f(t)}{n + 1}.
\end{align}
The proof follows by transforming it into the identity
\begin{align}
\sum_{j = 0}^{n} \sum_{k = j}^{n} (-1)^{k-j} \binom{k}{j} \binom{t}{k} f(j) = \sum_{k = 0}^{n} \binom{t}{k} (\Delta^{k} f)(0) = f(t),
\end{align}
where $\Delta^{k}$ is the $k^{\text{th}}$ forward difference operator. However, I'd like to prove the aforementioned identity directly, without recourse to the calculus of finite differences. Any hints are appreciated!
Thanks. 
 A: This is just Lagrange interpolation for the values $0, 1, \dots, n$.
This means that after cancelling the denominators on the left you can easily check that the equality holds for $t=0, \dots, n$.
A: I just ran across this question after working on this answer, and realized that the same method could be used here.
Notice that your equation is equivalent to
$$
\sum_{j=0}^n(-1)^{n-j}\binom{n}{j}\frac{f(j)}{t-j}=\frac{n!f(t)}{t(t-1)(t-2)\dots(t-n)}
$$
As long as $f$ is a polynomial of degree $n$ or less, apply the Heaviside Method for Partial Fractions to the right hand side to get the left hand side.
That is, to compute the coefficient of $\frac1{t-j}$ on the left hand side, multiply both sides by $t-j$ and set $t=j$. The right hand side becomes
$$
\frac{n!f(j)}{j(j-1)(j-2)\dots1(-1)(-2)\dots(j-n)}=(-1)^{n-j}\binom{n}{j}f(j)
$$
A: This can also be done with  complex variables. Observe that we have by
inspection that
$$\sum_{j=0}^n (-1)^j {n\choose j} \frac{f(j)}{t-j}
= (-1)^n \times n! \times 
\sum_{k=0}^n \mathrm{Res}_{z=k}
\frac{f(z)}{t-z}\prod_{q=0}^n \frac{1}{z-q}.$$
This holds  even if  $f(t)$ vanishes at  some positive integer  in the
range.

Recall that the residues sum to zero, so the right is equal to
$$-(-1)^n \times n! \times 
\sum_{k\in\{\infty,t\}} \mathrm{Res}_{z=k}
\frac{f(z)}{t-z}\prod_{q=0}^n \frac{1}{z-q}.$$
The residue at infinity of a function $h(z)$ is given by the formula
$$\mathrm{Res}_{z=\infty} h(z)
= \mathrm{Res}_{z=0} 
\left[-\frac{1}{z^2} h\left(\frac{1}{z}\right)\right]$$
which in the present case yields
$$-\mathrm{Res}_{z=0} \frac{1}{z^2}
\frac{f(1/z)}{t-1/z}\prod_{q=0}^n \frac{1}{1/z-q}
\\ = -\mathrm{Res}_{z=0} \frac{z^{n+1}}{z^2}
\frac{f(1/z)}{t-1/z}\prod_{q=0}^n \frac{1}{1-qz}
\\ = -\mathrm{Res}_{z=0} z^n
\frac{f(1/z)}{zt-1}\prod_{q=0}^n \frac{1}{1-qz}.$$
Note however that  $f(z)$ has degree $m$ and we  require that $n\ge m$
which means $f(1/z) z^n$ has no  pole at zero and hence the residue at
infinity is zero as well.

That leaves the residue at $z=t$ for a total contribution of
$$-(-1)^n \times n! \times (-1)\times f(t)\times
\prod_{q=0}^n \frac{1}{t-q}
= (-1)^n f(t) \times n! \prod_{q=0}^n \frac{1}{t-q}
\\ = (-1)^n f(t) \times n! \times 
\frac{1}{(n+1)!} {t\choose n+1}^{-1}
\\ = (-1)^n \frac{f(t)}{n+1} {t\choose n+1}^{-1}$$
which was to be shown, QED.
