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 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 $t=0, \dots, n$.
