Given $n\in\mathbb{N}$, define $\tau_n(g)=|\lbrace h\in G: h^n=g\rbrace|$.
Let $\chi_i,1\leq i\leq r$ be the distinct complex irreducible characters of a finite group $G$, and let $\gamma_n(\chi_i)=\frac{1}{|G|}\sum_{g\in G}\chi_i(g^n)$.
Prove that $\tau_n(g)=\sum_{i=1}^r\gamma_n(\chi_i)\chi_i(g)$.
I have been poking away at this for a couple days, and can't seem to get anywhere. I have manually checked that it works in groups like $Q_8$ and $S_4$, but I can't seem to wrap my head around why this is true, outside of the easy cases like $n = 1$ or where raising each element to the $n^{th}$ power induces an automorphism of sets, i.e $\lbrace g: g\in G\rbrace=\lbrace g^n:g\in G\rbrace$.
I'm most curious as to WHY this works. It seems so nice that it must have a simple solution hidden somewhere.
As has been discussed in the comments, the claim follows if we show $\langle \tau_n,\chi_i\rangle=\gamma_n(\chi_i)$. Things become a bit clearer if we rename the slightly misleadingly named summation variable $g$ to $h$:
$$\gamma_n(\chi_i)=\frac{1}{|G|}\sum_{h\in G}\chi_i(h^n)=\frac1{|G|}\sum_{g\in G}\tau_n(g)\chi_i(g)\;.$$