Solving Symmetric Group Induction Proof: Hints for Double Induction

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Homework Statement


Consider the symmetric group ##S_n##. I am trying to establish that ##(i,i+1)=(1,2,...,n)(i-1,i)(1,2,...,n)^{-1}##

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The Attempt at a Solution



I am trying to decide whether I need double induction or not. I have done several calculations to see whether I can get away with one induction, but it isn't clear to me whether this is possible. I could use some hints.
 
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Why don't you simply apply the mappings for the cases ##j< i-2\, , \,j > i-1## and ##j \in \{i-2,i-1\}## if read from left to right? It is no recursive definition but an explicit formula, so why use induction?
 
Hold on! I forgot that I proved that ##\sigma (x_1,x_2,...,x_n) \sigma^{-1} = (\sigma(x_1),...,\sigma(x_n))##, which makes the problem trivial.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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