Find the number of zeros of $f(z)=e^{z-1}-az$ inside unit disk, assuming $\mid a \mid >1$

Question

This is an application of Rouche's theorem, I want to make sure I am doing it correctly:

Let $f(z)=e^{z-1}-az$, where $\mid a \mid>1$ and $g(z)=-az$

Now, on the unit circle we have:

$$\mid g(z) \mid=\mid a \mid \mid z \mid=\mid a \mid >1$$

and

$$\mid g(z)-f(z)\mid=\mid e^{z-1} \mid=e^{Re(z-1)}=e^{Re(z)-1}=e^{Re(z)}e^{-1}\leq e^{-1}<1$$

Thus, on the unit circle $\mid g(z)-f(z) \mid< \mid g(z) \mid$. Therefore, $g$ and $f$ have the same number of zeros inside the unit circle, so $f(z)$ has one zero inside the unit circle.

Is this correct?