I'd like to ask how to prove or disprove the following identity (which I have to use on some computation regarding $\epsilon$-constant for constructing certain local Langlands parameter):


Let $\psi$ be any non-trivial additive character on $\mathbb{F}\_q$, the field of $q$ elements. Then


$$\sum\_{x\in\mathbb{F}\_{q^d}}\left(\frac{x}{q^d}\right)\psi(\text{Tr}\_{\mathbb{F}\_{q^d}/\mathbb{F}\_q}x)=\left(\sum\_{x\in\mathbb{F}\_{q}}\left(\frac{x}{q}\right)\psi(x)\right)^d,$$


where $\left(\frac{\cdot}{q^d}\right)$ denote the quadratic symbol on $\mathbb{F}\_{q^d}^\*$. The identity is definitely correct up to sign, and I come up with it just via numerical checking. Thanks!



Identity on Gauss sums
I think this is just the Davenport-Hasse Theorem, which [Wikipedia](http://en.wikipedia.org/wiki/Hasse%E2%80%93Davenport_relation) calls the Hasse-Davenport lifting relation. 


