Find bijection $\phi:[n]\times S_{n-1}\to S_n$ and conclude that $\vert S_n\vert = n!$ $[n] = \{1,...,n\}$
I am struggling with finding the bijection.
$\underline{\text{Assume } \mathbf{n=3:}}$
Than we have $A:=[n]\times S_{n-1} = \{(1,\pi_1), (2,\pi_1),(3,\pi_1),(1,\pi_2), (2,\pi_2),(3,\pi_2)\}$,
with $\pi_1 = id$ and $\pi_2 = (12)$
$S_3 = \{(123), (12)(3), (13)(2), (23)(1), (132), id\}$
So we are looking for some $\phi:A\to S_3$. The mapping for $a\in A \mapsto \sigma\in S_3$ is not obvious to me.
Has someone a hint?
The conclusion should be easy if the bijection is found.
Edit:
Due to your hints and further research I found
$\phi:[n]\times S_{n-1}\to S_n, (i,\pi_j)\mapsto \pi_j\circ (in)$, $i\in[n], j\in [n-1]$
My current status on $\phi^{-1}$ is:
$\phi^{-1}:S_n\to [n]\times S_{n-1}, \pi \mapsto (\pi^{-1}(n), ???)$
I am still struggling with $\phi^{-1}$. Does someone has a hint?

Well, we have $\pi_1 = \pmatrix{1 & 2\\ 1&2}$ and $\pi_2 = \pmatrix{1 & 2 \\ 2 & 1}$ or as sequences of the images $\pi_1 = 12$ and $\pi_2=21$ without delimiters.
Rule: in $(i,\pi)$ write $n$ in the $i$th position of $\pi$. Here:
$(1,p_1) = 312$, $(2,p_1)=132$ and $(3,\pi_1)= 123$ or $(1,\pi_1) = \pmatrix{1 & 2 & 3\\ 3&1&2}$, $(2,\pi_1) = \pmatrix{1 & 2 & 3\\ 1&3&2}$ and $(3,\pi_1) = \pmatrix{1 & 2 & 3\\ 1&2&3}$.
This demonstrates also  the general situation.
