$\underline{\textbf{Solution 1}}$

Applying $k\binom{n}{k} = n\binom{n-1}{k-1}$ gives $$\begin{align*} &\binom{n}{1} + 2\binom{n}{2}+ 3\binom{n}{3} + \cdots +n\binom{n}{n}\\ \\=\ &n\binom{n-1}{0}+n\binom{n-1}{1}+n\binom{n-1}{2}+\cdots + n\binom{n-1}{n-1}\\ \\=\ &n\left(\binom{n-1}{0}+\binom{n-1}{1}+\binom{n-1}{2}+\cdots + \binom{n-1}{n-1}\right)\\ \\=\ &n\cdot 2^{n-1}\end{align*}$$

$\underline{\textbf{Solution 2}}$

Adding a zero-value item at the beginning gives  $$\begin{align*} S=\ &\binom{n}{1}+ 2\binom{n}{2}+ \cdots +(n-1)\binom{n}{n-1}+n\binom{n}{n}\\ \\=\ &0\binom{n}{0} + 1\binom{n}{1}+ 2\binom{n}{2}+ \cdots +(n-1)\binom{n}{n-1}+n\binom{n}{n}\end{align*}$$

Reversing the order and also noting $\binom{n}{k}=\binom{n}{n-k}$ give $$\begin{align*} S =\ &n\binom{n}{n} + (n-1)\binom{n}{n-1}+\cdots + 2\binom{n}{2}+1\binom{n}{1}+0\binom{n}{0}\\ \\=\ &n\binom{n}{0}+ (n-1)\binom{n}{1}+ \cdots + (n-1)\binom{n}{n-1}+ n\binom{n}{n}\end{align*}$$

Adding these two relations yields $$\begin{align*} 2S =\ & n\displaystyle\binom{n}{0}+ n\binom{n}{1}+ n\binom{n}{2}+\cdots + n\binom{n}{n-1}+ n \binom{n}{n}\\ =\ & \displaystyle n\left(\binom{n}{0}+ \binom{n}{1}+ \binom{n}{2}+\cdots + \binom{n}{n-1}+ \binom{n}{n}\right)\\ \\=\ & n\cdot 2^n\end{align*}$$

It follows that $$S=n\cdot 2^{n-1}$$