Show $x^{p^n}-x$ has distinct roots over a field of characteristic $p$.

Well, $\dfrac{\mathrm d}{\mathrm dx} (x^m-x) = mx^{m-1}-1 = -1 \ne 0$, so the polynomial is separable, so the roots are distinct.

Bonus: the roots of $x^m - x$ actually form a field.

The usual way to show this is to define the formal derivative $f'$ of a polynomial $f$. For your polynomial $f$, the formal derivative is exactly what you think it would be by applying the power-rule on $f$. Then there's a theorem, the actual proof of which is not too bad but should be believable without proof from your experience in calculus, that goes

For a field $K$ let $F$ be the splitting field of $f \in K[x]$. Then $f$ has distinct roots in $F$ if $f$ and $f'$ have no common roots.

For your particular polynomial $x^{p^n} -x$, the formal derivative will be $p^n x^{p^n-1}-1$ which equals $-1$ since we're in a field of characteristic $p$, so the formal derivative doesn't even have roots, and we're good.

I think what you're attempting in your solution just unfolds the details of the proof of the above theorem.

You only have to show that $f(x)=x^{p^{n}}-x$ and $f'(x)$ have no common root, which results from the fact that in a field of characteristic $p$, $f'(x)=-1$.