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.