I am having trouble proving the identity below. $n$ and $k$ are positive integers.
$$ \left(\!\!{n\choose k}\!\!\right) = \sum_{i=0}^{k}\ \left(\!\!{n-1\choose i}\!\!\right)$$
I have tried multiple ways to relate it to something similar like the below to help get closer to something less complex, $$\sum_{i=0}^{k}\ \left(\!\!{n-1\choose i}\!\!\right) = {n+k -1\choose k}$$
however it doesn't seem to help.
One approach is to argue combinatorially. Suppose that you want to choose a $k$-element multiset from the set $[n]=\{1,\ldots,n\}$. Let $\mathscr{M}$ be the family of all such multisets, and for $\ell=0,\ldots,k$ let $\mathscr{M}_\ell$ be the set of multisets in $\mathscr{M}$ that have $\ell$ copies of $n$. If $M\in\mathscr{M}_\ell$, then what remains of $M$ when you throw out the copies of $n$ is a multiset of cardinality $k-\ell$ chosen from the set $[n-1]$, so
$$|\mathscr{M}_\ell|=\left(\!\!\binom{n-1}{k-\ell}\!\!\right)\;,$$
and of course
$$|\mathscr{M}|=\left(\!\!\binom{n}k\!\!\right)\;.$$
Can you finish it from here?
Another possibility is to reduce it to binomial coefficients and try to show that
$$\binom{n+k-1}k=\sum_{i=0}^k\binom{n-2+i}i\;.$$
This can be rewritten as
$$\sum_{i=0}^k\binom{n-2+i}{n-2}=\binom{n-1+k}{n-1}\;,$$
which is sometimes known as the hockey stick identity and has several proofs here.
