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. 

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.
