Let be $(G, \ast)$ a finite group, let's denote $|G|$ as the cardinality of $G$, i.e., the number of elements of $G$. Also, the family of all permutations of a set $U$, denoted by $S_U$, is called the symmetric group of $U$. When $U = \{1, 2, 3,\ldots,n\}, S_U$ is usually denoted by $S_n$ and it is called the symmetric group on $n$ letters.
I need to prove that $|S_n| = n!.$
Here's a hint. Let me know if more information is needed.
Let $\sigma\in S_U$ be a permutation. What is the number of possible values of $\sigma(1)?$ How about $\sigma(2)?$ Remember that $\sigma(1)\neq\sigma(2),$ since, by definition, $\sigma$ is injective. Now continue counting the possibilities for $\sigma(3),\sigma(4),\ldots,\sigma(n).$ What do you get? What does that tell you about the possible permutations on $U$?