Summit
Prove that the sum from t=0 to m of (-1)^t/t!(m-t)! is zero.
Prove that the sum
$$ \sum_{t=0}^m {(-1)^t\over t!(m-t)!} = 0 $$
Think about the binomial theorem.
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:
$$\sum_{t=0}^m \frac{(-1)^t}{t!(m-t)!}=0 $$