Pascal’s rule proof

We need to show

\\displaystyle\\binom{n}{k}+\\binom{n}{k-1}

\\displaystyle=

\\displaystyle\\binom{n+1}{k}

Let us begin by writing the left-hand side as

\\frac{n!}{k!(n-k)!}+\\frac{n!}{(k-1)!(n-(k-1))!}

Getting a common denominator and simplifying, we have

$\\displaystyle\\frac{n!}{k!(n-k)!}+\\frac{n!}{(k-1)!(n-k+1)!}$	$\\displaystyle=$	$\\displaystyle\\frac{(n-k+1)n!}{(n-k+1)k!(n-k)!}+\\frac{kn!}{k(k-1)!(n-k+1)!}$
	$\\displaystyle=$	$\\displaystyle\\frac{(n-k+1)n!+kn!}{k!(n-k+1)!}$
	$\\displaystyle=$	$\\displaystyle\\frac{(n+1)n!}{k!((n+1)-k)!}$
	$\\displaystyle=$	$\\displaystyle\\frac{(n+1)!}{k!((n+1)-k)!}$
	$\\displaystyle=$	$\\displaystyle\\binom{n+1}{k}$