TITLE: Does the identity $\det(I+g^{-1})\det(I+g)=|\det(g-I)|^2$ hold for $g \in U(n)$?
QUESTION [5 upvotes]: In a  paper (corollary 1, p.14) the following identity is used:

Let g be a unitary matrix. Then:
$$\det(I+g^{-1})\det(I+g)=|\det(g-I)|^2 \text{ for }g \in U(n)$$

Now my question is why this holds
I calculated:
$$\det(I+g^{-1})\det(I+g)=\overline{\det(I+g^t)}\det(I+g)=\overline{\det(I+g)}\det(I+g)=|\det(I+g)|^2$$
Where the second equality holds as $I$ has only entries in the diagonal ($I$ is of course the unit matrix). But this is not the same as on the right side.
(I also thought that maybe there was a typo on the left side where should be minus-signs. However in the paper itself it is needed that there are plus-signs.)
Thanks for any hints.
Edit:
This equality was in the scope of an integral: $$\int_{U(n)}\prod_{l=1}^{k}det(I+g^{-1})\prod_{l=1}^{k}\det(I+g)dg=\int_{U(n)}|\det(g-I)|^{2k}dg$$
With a change of variable it was solved with my calculation done above. See Giuseppe's answer.

REPLY [6 votes]: Dear AndreasS I have given a look at the paper.  
I think that the calculation  $$\det(I+g^{-1})\det(I+g)=|\det(I+g)|^2 \text{ for }g \in U(n)$$ is correct.
But in order to obtain the result stated in Corollary 1, you just need the change of variable $g\mapsto -g$ in the integral over $U(n)$, so that $$\int_{U(n)}|\det(g-I)|^{2k}dg=\int_{U(n)}|\det(I+g)|^{2k}dg.$$
Then in the paper you find how factorize $|\det(I+g)|^{2k}.$
I hope that it helps.