Use the identity $(xy)^n = x(yx)^{n-1}y$ to prove that $xy$ and $yx$ always have the same spectral radius. Both $x,y$ belongs to a Banach Algebra.
My attempt:
$(\lambda e - xy) = \sum_{j=0}^n (\lambda e)^j (xy)^{n-j}$ and then I tried to put $(xy)^{n-j} = x (yx)^{n-j-1} y$ and then try to spot something useful, but I'm not sure if it was the correct thing to do or not...