This is a General Linear Models topic but I believe it's just basic failure to remember some more basic math rules that's making it difficult for me.
If the link function is
$$g(\pi) = \log(\frac{\pi}{1-\pi}) = x^T\beta$$
show that this is equivalent to modeling the probability $\pi$ as,
$$\pi = \frac{e^{x^T}\beta}{1+e^{x^T}\beta}$$
Again I think this is a simply inverse. But I'm not quite sure how to do it.
You're right, it is just the inverse
$$ \ln\frac{\pi}{1 - \pi} = x^T\beta ~~~\Rightarrow~~~ \frac{\pi}{1 - \pi} = e^{x^T\beta} ~~~\Rightarrow~~~ \pi = (1 - \pi)e^{x^T\beta} = e^{x^T\beta} - \pi e^{x^T\beta} $$
Now rearrange a bit things
$$ \pi (1 + e^{x^T\beta}) = e^{x^T\beta} $$
and from here
$$ \pi = \frac{e^{x^T\beta}}{1 + e^{x^T\beta}} $$