I recently discovered this identity:
For all natural numbers n, $$\sum_{k=1}^n\left(\prod_{i=1}^k\frac{n-k+i}{k-i+1}\right)=2^n-1$$
I already know one way to prove it, since I derived the identity myself using the following two facts:
$$\sum_{k=1}^n {n\choose k} = 2^n - 1\tag{1}$$ $${n\choose {r+1}} = {n\choose r}\times\frac{n-r}{r+1}\tag{2}$$
A generalization of equation $(2)$ for ${n\choose {r+k}}$ is: $${n\choose {r+k}} = {n\choose r}\prod_{i=1}^k\frac{n-r-k+i}{r+k-i+1}\tag{3}$$
Setting $r=0$ in equation $(3)$, we obtain:
$$\begin{align} {n\choose k} &= {n\choose 0}\prod_{i=1}^k\frac{n-0-k+i}{0+k-i+1} \\ &=\prod_{i=1}^k\frac{n-k+i}{k-i+1} \end{align}$$
Substituting the above expression for ${n\choose k}$ into equation $(1)$ gives us the identity. My question is: Are there alternate ways to prove this identity?