Finding the upper bound of a complex contour integral

I am trying to show that $$\left\lvert\int_\gamma \frac{\cos(z)}z \,dz\right\rvert \le 2\pi e$$ if $$\gamma$$ is a path that traces the unit circle once.

The book recommends that I show that $$\lvert \cos(z) \rvert \le e$$ if $$\lvert z \rvert = 1$$ to help prove this. I know that if $$\lvert f(z) \rvert \le M$$ for all $$z \in \gamma (I)$$ then $$\left\lvert \int_\gamma f(z) \,dz \right\rvert \le M\ell (\gamma),$$ where $$\ell (\gamma)$$ is the length of the path, which in this case is $$2\pi$$. So I can see why I would need to show $$\lvert cos(z) \rvert \le e$$ if $$\lvert z \rvert = 1$$ to prove the inequality, but I am not sure where to go from here to show that.

With the series expansion of $$\cos$$ we get for $$|z|=1$$:
$$| \cos z| \le \sum_{n=0}^{\infty}\frac{1}{(2n)!} \le \sum_{n=0}^{\infty}\frac{1}{n!}=e.$$
The exponential function satisfies the inequality $$\vert e^z \vert \leq e^{\vert z \vert}$$. We have $$\cos(z) = \frac{1}{2}(e^{iz} + e^{-iz}),$$ thus \begin{equation*} \begin{aligned} \vert \cos(z) \vert &\leq \frac{1}{2}( \vert e^{iz} \vert + \vert e^{-iz} \vert) \\ &\leq \frac{1}{2}(e^{\vert z \vert} + e^{\vert z \vert}) \\ &= e^{\vert z \vert}. \end{aligned} \end{equation*} So when $$\vert z \vert = 1$$ we have $$\vert \cos(z) \vert \leq e$$.
$$|\cos z|=|\frac {e^{iz}+e^{-z}} 2| \leq \frac {e+e} 2=e$$ because $$|e^{z}| \leq e^{|z|}$$.