Proof of $k {n\choose k} = n {n-1 \choose k-1}$ using direct proof I've seen many posts regarding a combinatorial proof of the following question. But for a non-combinatorial proof would the following method work? Also... is this the easiest way to arrive at a proof? It seems to be rather verbose.
Show the formula $k {n\choose k} = n {n-1 \choose k-1}$ is true for all integers $n,k$ with $0\le k \le n$.
My answer:
Observe that
$$\begin{align*}
{n\choose k}&= \frac {n}{k} {n-1 \choose k-1}\\
&= \frac {n}{k}\left[ {n-2 \choose k-2} + {n-2 \choose k-1}\right]\\
&= \frac {n}{k}\left[ \frac{(n-2)!}{(k-2)!(n-2-(k-2))!} + \frac {(n-2)!}{(k-1)!(n-2-(k-1))!}\right]\\
&= \frac {n}{k}\left[ \frac{(n-2)!}{(k-2)!(n-k)!} + \frac {(n-2)!}{(k-1)!(n-k-1)!}\right]\\
&= \frac{(n-1)!}{(k-1)!(n-k)!} + \frac {(n-1)!}{(k)!(n-k-1)!}\\
&= \frac{(n-1)!}{(k-1)!(n-1-(k-1))!} + \frac {(n-1)!}{(k)!(n-1-k)!}\\
&={n-1 \choose k-1} + {n-1 \choose k}={n \choose k}.
\end{align*}$$

Note that you have
$$n\binom{n-1}{k-1}=n\frac{(n-1)!}{(n-k)!(k-1)!}=\frac{n!}{(n-k)!(k-1)!}$$ and $$k\binom{n}{k}=k\frac{n!}{(n-k)!k!}=\frac{n!}{(n-k)!(k-1)!}.$$
