Permutation and Group

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A permutation of a permutation, is the product of the two permutations?
Right. The multiplication is not commutative, i.e in general ab≠ba. For example $$\begin{pmatrix}1&2&3\\1&3&2\end{pmatrix}\begin{pmatrix}1&2&3\\2&3&1\end{pmatrix}=\begin{pmatrix}1&2&3\\3&2&1\end{pmatrix}$$ but $$\begin{pmatrix}1&2&3\\2&3&1\end{pmatrix}\begin{pmatrix}1&2&3\\1&3&2\end{pmatrix}=\begin{pmatrix}1&2&3\\2&1&3\end{pmatrix}.$$
What happens if I keep permuting using the same permutation?
If you keep permuting with same permutation, you will eventually return to the initial configuration. For example $$\begin{pmatrix}1&2&3\\2&3&1\end{pmatrix}\begin{pmatrix}1&2&3\\2&3&1\end{pmatrix}\begin{pmatrix}1&2&3\\2&3&1\end{pmatrix} =\begin{pmatrix}1&2&3\\2&3&1\end{pmatrix}\begin{pmatrix}1&2&3\\3&1&2\end{pmatrix} =\begin{pmatrix}1&2&3\\1&2&3\end{pmatrix}.$$ The identity permutation, usually denoted by (1), is the permutation that changes nothing. We know that, say, there are n objects, then applying the same permutation for n! times will always end up to be the identity permuation, i.e. $$\tau^{n!}=(1)$$
Can I say τ^n!-1=1/τ?
Not quite. But you can write τ^n!-1=τ^-1, meaning that it is the multiplicative inverse of τ in the permutation multiplication.

Muliplication with an identity element and inverses. It is a group!?
Right. It is a group.
For every element α of a group G, we define a function A_α(g)=αg. It is not hard to see that A_α is a bijection and so it is a permutation. Indeed we have A_αA_β=A_αβ. Hence if we identify α with A_α, we see that each element of a group is indeed a permutation of the group!