This neat solution came from Marcos and the result was also
proved by Yatir Halevi:

By the binomial expansion:

$$(1+x)^m=\sum_{t=0}^m \frac{m!}{t!(m-t)!}x^t $$

This can be proved by induction on $m$ but I won't clutter
this with unnecessary proofs.

Putting in $x= -1$ we have

$$0=\sum_{t=0}^m \frac{m!}{t!(m-t)!}(-1)^t
$$

Dividing through by $m!$ gives us the required
result:

$$\sum_{t=0}^m \frac{(-1)^t}{t!(m-t)!}=0 $$