Binomial coefficient as a summation series proof?

Alright, so I was wondering if the following is a well known identity or if its existence provides any real benefits other than serving as a time-saver when dealing with higher values for combinations.

After screwing around with some basic combinations stuff, I noticed the following:

$$\sum_{i=1}^{n-1} i = \begin{pmatrix}n\\2\\ \end{pmatrix}$$

To prove this, I used Gauss' method to simplify the summation, and I wrote n choose 2 in terms of factorials to simplify the right side.

$$\frac{ (n-1) n } {2} = \frac{ n! } { (2!) (n-2)! }$$

$$2!(n-2)!(n-1)(n) = 2n!$$

$$2(n-2)!(n-1)(n) = 2n!$$

$$(n-2)!(n-1)(n) = n!$$

$$n! = n!$$

I did this on lunch break one day over the summer. I'm in high school, so my math skills are very subpar on this forum, but I was hoping some people might discuss it and/or answer my aforementioned questions. I didn't see anything about it on here or Google, for that matter. If you found this banal or rudimentary, just let me know and I'll refrain from posting until I come up with something more interesting. Regardless, I hope you found it worth your time.

Answer

I was wondering if the following is a well known identity

Not only is it well-known, but it's part of a much larger group. In general, we have

$$\sum_{k=0}^nk~(k+1)~\cdots~(k+p)~=~(p+1)!~{n+p+1\choose n-1}~=~(p+1)!~{n+p+1\choose p+2}$$

The whole idea is to rewrite the summand as $(p+1)!~\displaystyle{p+k\choose p+1}.~$ See also Faulhaber's formulas.