# Perform index shift on summation containing factorial

I'm having a hilariously hard time solving a problem that looks/feels so easy but just won't open up to me. I'm trying to show this equality:

$$\sum_{k=0}^{n-1} \frac{1}{k!(n-k-1)!} = \sum_{k=1}^{n} \frac{1}{(k-1)!(n-k)!}$$

It's so very obvious that the results in these sums are always the same, just that k is shifted by 1 on the right side, but I cannot figure out how extract the right terms to shift the indices properly.

Would really appreciate some help.

It is useful to make some intermediate passages using another variable. Here we use $j=k+1$; every time that $k$ appears, we write $k=j-1$ instead.
$$\sum_{k=0}^{(k=)n-1} \frac{1}{k! (n-k-1)!} = \sum_{j-1=0}^{(j-1=)n-1} \frac{1}{(j-1)! (n-(j-1)-1)!} = \sum_{j=1}^{(j=)n} \frac{1}{(j-1)! (n-j+1-1)!} = \sum_{j=1}^{n} \frac{1}{(j-1)! (n-j)!}.$$
After doing this, you change again the variable, going back to $k$ by putting $k=j$:
$$\sum_{k=0}^{n-1} \frac{1}{k! (n-k-1)!} = \sum_{j=1}^{n} \frac{1}{(j-1)! (n-j)!} = \sum_{k=1}^{n} \frac{1}{(k-1)! (n-k)!} .$$