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. 

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
$$
\frac{n!f(j)}{j(j-1)(j-2)\dots1(-1)(-2)\dots(j-n)}=(-1)^{n-j}\binom{n}{j}f(j)
$$
