Proving $\left(\binom{n}{k}\right)=\left(\binom{n-1}{0}\right)+\left(\binom{n-1}{1}\right)+\cdots+\left(\binom{n-1}{k}\right)$

Question

Here, $\left(\binom{n}{k}\right)$ denotes the number of multisets in $N$ with length $k$.

I can prove it using the fact that $\left(\binom{n}{k}\right) = \binom{n+k-1}{k}$ but I want another access.

Please help.

Answer 1

You can have between $0$ to $k$ occurrences of the $n$-th element, and the rest of the multiset is a multiset of the other $n-1$ elements with $k$ to $0$ elements.

Answer 2

How about a proof using combinatorial argument? In how many ways can you choose multisets from a total of $n$ elements with length $k$ where $0 \le k \le n$?

HINT: While vying for the R.H.S., consider that the $n$-th element can occur $0$ to $k$ times while the rest is a multiset of $0$ to $k$ elements.