Showing that $f$ has exactly one fixed point

Let $\gamma$ be the circle $\{z \in \mathbb{C}: \lvert z\rvert=1 \}$. Suppose $f$ is a function analytic on an open set containing $\gamma$ and its interior and that $\lvert\, f(z)\rvert<1$ for each $z$ on $\gamma$. Show that $f$ has exactly one fixed point inside $\gamma$

That is, there is exactly one $z$ in the open unit disk with $f(z)=z$.

Is this a result of Louville's theorem?

I don't know how to approach it.

Let $g(z)=-z$ and $h(z)=f(z)-z$.
Since $$\lvert h(z)-g(z)\rvert=\lvert\, f(z)\rvert <1=\lvert g(z)\rvert,$$ for every $z$ on $\gamma$, then the functions $g$ and $h$ have the same number of zeros in the interior of $\gamma$. Clearly $g$ has exactly one zero, and hence $f$ has exactly one fixed point inside the unit circle.