Combinatorial argument for $\sum\limits_{k=i}^{n}\binom{n}{k}\binom{k}{i} = \binom{n}{i}2^{n-i}$

HINT: On both sides you’re actually dividing the pool of $n$ things into three sets. One of those sets contains exactly $i$ objects, one contains $k-i$ for some $k$, and the third contains the $n-k$ that are left. For the righthand side, note that once you’ve chosen two of those sets, you no longer have to choose the third: it’s simply what’s left.

(This is a skimpier hint than I often give, because you’ve already done most of the work: you just need to see how to put the pieces together.)