Suppose that $|A|=n$ and that $\Sigma_A$ be the set of permutations of $A$, then determine the cardinality of $\Sigma_A$.
I found that the reasoning below is not rigorous and convincing enough for me to trust since it is not built up on any mathematical principle. Maybe I have not learned anything in combinatorics and just exposed to some elementary set theory, so I'm unable to understand the underlying theorems/principles behind such everyday words as choose.
Since $|A|=n$, then assume $A=\{a_1,a_2,\ldots,a_n\}$. Let $f:A\to A$ be a bijection, and thus $f$ is a permutation of $A$. We have $n$ ways to choose an element for $f(a_1)$, $n-1$ ways to choose an element for $f(a_2)$, and so on. Thus we have $n\cdot(n-1)\cdots1=n!$ ways to generate $f$. Thus $\Sigma_A=n!$.
I also found that the proof is quite straightforward by induction on $n$ if I know $|\Sigma_A|=n!$ in advance, but I can not imagine any way to figure out the intuition behind $|\Sigma_A|=n!$. I'm sure that I would be beaten if I did not know the result.
My questions:
Is there any intuition behind the formula $|\Sigma_A|=n!$.
If I did not know $|\Sigma_A|=n!$ in advance, how would I visualize/sketch a diagram to determine $|\Sigma_A|$?
The common way to conjecturize a result you have no hint about is to check what happens for small values of $n$.
It takes only a few seconds to find out that $|\Sigma_{A_1}|=1$ (where $|A_n|=n$), $|\Sigma_{A_2}|=2$ and $|\Sigma_{A_3}|=6$.
A little effort for listing all the permutations of $A_4$ might take two minutes; Alternatively you can reason by noticing that a permutation of $\{1,2,3,4\}$ is a permutation of $\{1,2,3\}$ with a $4$ added at any of $4$ possible spots (and you are already on your way to the proof by induction).
Once you get $|\Sigma_{A_4}|=24$, you should recognize the !-pattern.