Question regarding algebraic proof for Pascal's identity. I was looking at the proof for
$\binom{n}{k}=\binom{n-1}{k}+\binom{n-1}{k-1}$, where $n$ and $k$ are each $\ge 1$.
According to the proof, expressing the right-hand side in terms of factorials, we get
$$\frac{(n-1)!}{k!(n-k-1)!}+\frac{(n-1)!}{(k-1)!(n-k)!}.$$
However, I don't understand how we got $(n-k)!$ instead of $(n-k-2)!$ for the denominator in 2nd term.
The equality used is the following:
$\binom{n}{k}=\frac{n!}{k!(n-k)!}$
So shouldn't we have $(k-1)!(n-k-2)!$ in the denominator for the second term?
I assume I'm missing something really simple, but any help would be appreciated.

No, it shouldn't be $n-k-2$, and, yes, you're missing something simple.
When you subtract $k-1$ from $n-1$, don't forget to distribute the $-1$ factor properly:
$(n-1)\color{blue}-(k\color{blue}-1)=n-1\color{blue}-k\color{blue}+1=n-k$.
