Prove formula equivalence without induction

Question

Consider the following formula:

$$\frac{(n-i)!}{n!}\sum_k \frac{(k-1)!}{(k-i)!} [2 \le i \le k \le n] = \frac{1}{i}$$

It's easy to prove the equality by induction over $n$ but is there also a more direct way (a continued equality, ideally)?

Answer 1

OPs identity is the Hockey stick identity in disguise.

We write OPs identity as \begin{align*} \frac{(n-i)!}{n!}\sum_{k=i}^n\frac{(k-1)!}{(k-i)!}=\frac{1}{i}\qquad\qquad 2\leq i\leq n\tag{1} \end{align*} Multiplying both sides of (1) with $\frac{n!}{(n-i)!(i-1)!}$ gives \begin{align*} \sum_{k=i}^n\binom{k-1}{i-1}=\binom{n}{i}\qquad\qquad 2\leq i\leq n\tag{2} \end{align*}

Shifting the index $k$ by one results in \begin{align*} \color{blue}{\sum_{k=i-1}^{n-1}\binom{k}{i-1}=\binom{n}{i}\qquad\qquad 2\leq i\leq n} \end{align*} which is the Hockey stick identity.