One of my favorite facts from combinatorics is that $\sum\limits_{0\leq k\leq n}\binom{n}{k} = 2^{n}$. To prove it is simple: Note that $2 = 1 + 1$, and that $\binom{n}{k} = \binom{n}{k}\cdot 1^{k}\cdot 1^{n-k}$, and appeal to binomial theorem:
$$\sum_{0\leq k\leq n}\binom{n}{k} = \sum_{0\leq k\leq n}\binom{n}{k}\cdot 1^{n-k}\cdot 1^{k} = \left(1+1\right)^{n} = 2^{n}.$$
But what can we say about $\sum\limits_{1\leq k\leq n}k\binom{n}{k}$, or maybe even more general binomial coefficient sums?