How to prove this formula?

Question

For every integer $n > 1$:
$$2^n = 2 + \sum_{x=1}^{n-1}\frac{ n!}{(n-x)!x!}$$
I found this equality while doing some probability stuff... Basically both formulas equal the number of possible combinations of $n$ coin tosses. Any other way of showing this to be true?

Answer 1

Yes: binomial formula:\begin{align}2^n&=(1+1)^n\\&=\sum_{x=0}^n\binom n{x}\\&=2+\sum_{x=1}^{n-1}\binom nx.\end{align}

Answer 2

Most popular choices to prove $$2^n=2+\sum_{x=1}^{n-1} \frac{n!}{x!(n-x)!}$$ are induction or the binomial theorem: \begin{align} (a+b)^n&=\sum_{x=0}^{n} \frac{n!}{x!(n-x)!} a^{n-x} b^x;\\ 2^n=(1+1)^n&=\sum_{x=0}^{n} \frac{n!}{x!(n-x)!} =2+\sum_{x=1}^{n-1} \frac{n!}{x!(n-x)!}. \end{align}