Let $n>1$.
Let $\tau \in S_{n+1}$ and $1≦l≦n+1$.
Assume that $\tau(l)=l$.
Now, define a permutation $\mu\in S_n$ as $\mu(i)=\tau(i)$ if $i<l$ and $\mu(i)=\tau(i+1)$ if $l≦i≦n$. (If $l=n+1$, then define $\mu$ as the restricton of $\tau$ on $\{1,...,n\}$.)
How do I show that $sgn(\mu)=sgn(\tau)$?
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Below is what I have proved, so you can use these if you need:
$sgn(\tau\sigma)=sgn(\tau)sgn(\sigma)$
$sgn(\tau)=\prod_{1≦i<j≦n}\frac{\tau(j)-\tau(i)}{j-i}$
There are a few equivalent definitions of the signature of a permutation. One of them is $$sgn(\tau)=(-1)^k,$$where $\tau$ can be represented as the product of $k$ transpositions. In the above case, given a representation of $\tau$ as a product of transpositions, $$\tau=t_1\cdot\ldots\cdot t_k,$$we may assume that each one of these transpositions fixes $l$, and so it is easy to adjust these transpositions and represent $\mu$ as $$\mu=t'_1\cdot\ldots\cdot t'_k,$$thus both permutations have the same signature.
As a matter of fact, this is only a consequence of a more general claim. Set$$P_l:=\{\sigma\in S_{n+1}|\sigma(l)=l\},$$ then $P_l$ is obviously a subgroup, and the map $\tau\mapsto\mu$ which is described in the question is an isomorphism of $P_l$ onto $S_n$.