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
$$\begin{align*}
{n\choose k}&= \frac {n}{k} {n-1 \choose k-1}\\
&= \frac {n}{k}\left[ {n-2 \choose k-2} + {n-2 \choose k-1}\right]\\
&= \frac {n}{k}\left[ \frac{(n-2)!}{(k-2)!(n-2-(k-2))!} + \frac {(n-2)!}{(k-1)!(n-2-(k-1))!}\right]\\
&= \frac {n}{k}\left[ \frac{(n-2)!}{(k-2)!(n-k)!} + \frac {(n-2)!}{(k-1)!(n-k-1)!}\right]\\
&= \frac{(n-1)!}{(k-1)!(n-k)!} + \frac {(n-1)!}{(k)!(n-k-1)!}\\
&= \frac{(n-1)!}{(k-1)!(n-1-(k-1))!} + \frac {(n-1)!}{(k)!(n-1-k)!}\\
&={n-1 \choose k-1} + {n-1 \choose k}={n \choose k}.
\end{align*}$$

The most straightforward proof uses analysis.
Remember the binomial coefficient  $\dbinom nk$ is usually defined as the coefficient as the coefficient of $x^k$ in the expansion of $(1+x)^n$  as a product of $n$ factors:
$$ (1+x)^n=\sum_{k=0}^n\binom nk x^k. $$
Differentiate this relation:
$$n(1+x)^{n-1}=\begin{cases}\displaystyle \sum_{k=0}^n k\binom nk x^{k-1}\\\displaystyle n\sum_{k=0}^{n-1}\binom{n-1}k x^k\end{cases}
$$
Identify the coefficients of $x^{k-1}$ for $k=1,\dots,n$. You obtain
$$k\binom nk =n\binom{n-1}k.$$
