I would like to give another different proof of the problem the OP proposed. My solution is based on the identity
$k\binom{n}{k}=n\binom{n-1}{k-1}.$
Let us first prove this identity: suppose we are given a class of $n$ children and suppose we want to form a team of $k$ people from the class, and moreover we want to elect a captain for our team. We can count the possibilities of doing so in two ways:
First select $k$ people from the class and then elect the captain. Then we have $k$ possibilities for any previously chosen team, so in total $k\binom{n}{k}$ ways of proceeding along this path.
But we may also elect first the captain, which can be done in $n$ ways, then form the team, for which we need other $k-1$ children out of $n-1$ remaining. In this other way we count $n\binom{n-1}{k-1}$ ways to fulfill our task.
This proves in a combinatorical way the identity which can be however verified by algebraic means.
But then our formula reduces to $ n \sum_{k=0}^{n} (-1)^k\binom{n-1}{k-1}=n\sum_{k=1}^n(-1)^k\binom{n-1}{k-1}=0.$