(1) Why is (r+1)r!=(r+1)!
(2)This is commonly asked : why is 0! =1 ??
$\displaystyle (r+1)r! = (r+1) \times r(r-1)(r-2)(r-3)\times ...\times 1 $
$\displaystyle (r+1)! = (r+1)(r+1-1)(r+1-2)(r+1-3)\times ...\times 1 = (r+1)(r)(r-1)(r-2)\times ...\times 1 $
See how they are identical?
Also. We know that $\displaystyle n!= n(n-1)(n-2)(n-3)...1 $. This can be written $\displaystyle n! = n(n-1)! $
So, let's look at n = 1, knowing that $\displaystyle 1! = 1 $
$\displaystyle 1! = 1(1-1)! $
$\displaystyle 1 = 1(0)! $
$\displaystyle 1 = 0! $.