Prove that: $\sum_{n_1+n_2+n_3 = n} \binom{n}{n_1,n_2,n_3} \cdot (-1)^{n_2} = 1$
The sum is over all positive integer solutions of $n_1 + n_2 + n_3 = n$.
Use the multinomial theorem:
$(x_1 + x_2 + x_3)^n = \sum_{n_1 + n_2 + n_3 = n} \binom{n}{n_1,n_2,n_3} x_1^{n_1} x_2^{n_2} x_3^{n_3}$
Let $x_2 = z$, $x_1 = x_3 = 0$ and plug into the above:
$(0 + z + 0)^n = \sum_{n_1 + n_2 + n_3 = n} \binom{n}{n_1,n_2,n_3} 0^{n_1} z^{n_2} 0^{n_3} = \sum_{n_1 + n_2 + n_3 = n} \binom{n}{n_1,n_2,n_3} z^{n_2}$
Let $z = -1$ and plug in:
$\sum_{n_1 + n_2 + n_3 = n} \binom{n}{n_1,n_2,n_3} (-1)^{n_2}$
But can't $n_2$ be both even and odd, so it could be either $1$ or $-1$?