What is the factorial of (n+1) ?

It is ${\displaystyle n!\cdot (n+1)}$
Explanation:
Since factorial n (or n!)) is the product of all numbers up to and including n, we only have to multiply by the next number.

As we know ${\displaystyle n\ne n\times (n-1)!}$
Put ${\displaystyle n=n+1}$ in above equation, hence
${\displaystyle (n+1)\ne (n+1)\times [(n+1)-1]!}$
${\displaystyle (n+1)=(n\times 1)\times [n+1-1]!}$
${\displaystyle (n+1)\ne (n+1)\times n}$
Hence proved.

What is the factorial of n?
If n is some positive integer, then the factorial of n is the product of every natural number till n, or
$n!=1\cdot 2\cdot 3\cdot ...\cdot n$
And that way, the factorial of n+1 becomes
$(n+1)!=1\cdot 2\cdot 3\cdot ...\cdot n\cdot (n+1)$
As you can clearly observe, the part of the second expansion till n is equal to the first expansion, the one for n factorial.
So,
$(n+1)!=n!\cdot (n+1)$