<s>[INST] Given $x^{p^2} = 1, x^p = y^p, yxy^{-1}=x^{p+1}$, show that $(yx^{-1})^p=1$ 
Suppose $p$ is an odd prime. Given $x^{p^2} = 1, x^p = y^p, yxy^{-1}=x^{p+1}$, show that $(yx^{-1})^p=1$, where $1$ is the identity.

I know that the conjugate of power is the power of conjugate, so $yx^{k}y^{-1} = (x^{p+1})^{k}$, and used this to obtain $yx^{-1} = x^{-p-1} y$, which will allow me to switch $y$ and $x^{-1}$ in each $yx^{-1}$. However, I still cannot figure out where to go from here. Any help is appreciated.
 [/INST] $yx^k = x^{k(p+1)}y$, so $y^a x^k = x^{k(p+1)^a} y^a$, so:
$$\begin{array}{rcl}
(yx^{-1})^p
&=& y x^{-1} y x^{-1} y x^{-1} \cdots yx^{-1} \\
&=& x^{-(p+1)}y^2 x^{-1} y x^{-1} \cdots yx^{-1} \\
&=& x^{-(p+1)-(p+1)^2} y^3 x^{-1} \cdots yx^{-1} \\
&=& \cdots \\
&=& x^{-(p+1)-(p+1)^2-\cdots-(p+1)^{p-1}} y^p \\
&=& x^{-(p+1)-(2p+1)-\cdots-((p-1)p+1)} x^p \\
&=& x^{-\frac{(p-1)p}2 p - p} x^p \\
&=& x^{-\frac{(p-1)}2 p^2} \\
&=& 1 \\
\end{array}$$
where $x^{(p+1)^n} = x^{1+np+O(p^2)} = x^{1+np}$ because $x^{p^2}=1$.
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