This problem can be solved by applying the hockey stick identity. $$\begin{align*} &\binom{n}{k} + 2\binom{n-1}{k}+ 3\binom{n-2}{k}+ \cdots + (n-k+1)\binom{k}{k}\\ \\=\ &\left(\binom{n}{k}+ \binom{n-1}{k}+ \binom{n-2}{k}+ \cdots + \binom{k}{k}\right) + \\ &\left(\binom{n-1}{k}+ \binom{n-2}{k}+ \cdots +\binom{k}{k}\right) + \\ &\left(\binom{n-2}{k}+ \cdots + \binom{k}{k}\right)+\\ & \cdots \\ & \binom{k}{k}\\ \\=\ &\binom{n+1}{k+1} + \binom{n}{k+1} + \binom{n-1}{k+1} +\cdots + \binom{k+1}{k+1}\\ \\=\ & \boxed{\binom{n+2}{k+2}} \end{align*}$$