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.