I'll delete this if somebody puts an answer that isn't quite as lazy as this. But for now, here's a hint:
What would you expect, in general, to be the eigenvalues of a matrix multiplied by its (conjugate) transpose? If for an orthogonal matrix, doing this gives you the identity (which has only 1 as its eigenvalues), what does that tell you?
EDIT: some elaboration
Okay, so a proof that the eigenvalues have magnitude 1:
We note that for any orthogonal matrix M, we have
<x,y> = <M^t M x, y> = (M^t M x)^t y = (M x)^t M y = <M x , M y>.
That is, all orthogonal matrices preserve the dot product of two vectors.
It follows that
||Mx||^2 = <M x, M x> = <x, x> = ||x||^2
That is, Mx has the same magnitude as x.
If v is an eigenvector with eigenvalue L, then Mv = L*v. However, we have
||v|| = ||M v|| = ||L * v|| = |L| * ||v||
That is, we deduce that ||v|| = |L| * ||v||, from which we conclude that |L| = 1.
It follows that all eigenvalues have magnitude 1.
Now, if M doesn't have complex eigenvalues, it means that M is either a reflection or the identity. If M has complex eigenvalues, then M is not purely a reflection or the identity. For example, any rotation matrix has complex eigenvalues.
Hope that helps