Show that norm of matrix $A$ is given by the square root of the largest eigenvalue of $A^tA$

The norm is defined as $\|A\|=\sup\{ \|A v \| : \|v\|=1\}$. I want to show it is equal to the square root of the largest eigenvalue of $A^tA$.

I do not know why it is an eigenvalue of a product of $A^tA$ not simply an eigenvalue of $A$. How to proceed?

The singular value decomposition of $A$ gives us orthogonal matrices $U, V$ and a diagonal matrix $S$ such that

$$A = U S V^T$$

Since $U$ and $ V$ are orthogonal, we have $\|U\| = \|V\| = 1$. Therefore, for all vectors $x$, we have

$$\|A x\| = \|S x\|$$

Since $S$ is the diagonal matrix containing the singular values of $A$ (which by definition are the roots of the eigenvalues of $A^T A$), the $x'$ which maximizes $\|S x\|$ is the unit vector $e_1 = (1, 0, \dots, 0)$ assuming the singular values in $S$ are sorted in descending order of magnitude.

Let $s$ be the largest singular value, then we have

$$\|A\| = \|A x'\| = \|S x'\| = \|S e_1\| = \|e_1 s\| = s$$

So, the norm of $A$ is indeed the largest singular value $s$ of $A$, which is the root of the largest eigenvalue of $A^T A$.