How can I finish my proof that $||A^*A||=||A||^2$? 
Let $A:X\rightarrow X$ be a bounded linear operator on a Hilbert space $X$. I want to show that $||A^*A||=||A||^2$. My idea was the following:

Let me consider $x\in X$ such that $||x||\leq 1$, then by the Cauchy-Schwarz inequality$$\begin{align}0\leq ||Ax||^2&=\langle Ax,Ax\rangle\\&=\langle x, A^*Ax\rangle\\&\leq ||x||\cdot ||A^*Ax||\\&\leq||A^*Ax||\end{align}$$ so by definition it follows $||A||^2\leq ||A^*A||.$On the other hand we know that $$\begin{align}||A^*A||&\leq ||A^*||\cdot ||A||\\&\stackrel{*}{\leq}||A||^2\end{align}$$Here in $*$ I'm not completely sure, I know that $A^*=J_x^{-1}A'J_X$, where $$J_X:X\rightarrow X':=\{f:X\rightarrow \Bbb{C}: f~~\text{linear functional}\};~~x\mapsto \langle\cdot, x\rangle$$is a norm preserving bijection, and where $$A':X'\rightarrow X';~~~f\mapsto f\circ A$$But then if we pick $x\in X$ such that $||x||\leq 1$ $$||A^*x||=||J_X^{-1}A'J_X x||\stackrel{\text{norm preserving}}{=}||A'J_X x||...$$ But here I don't see how to continue with this norm, could maybe someone help me how to continue from there? I think I'm not far away
 A: Hint. Using the fact that
$$
\|B\|=\sup_{\|x\|=\|y\|=1}|\langle Bx,y\rangle|
$$
we obtain
$$
\|A^*A\|=\sup_{\|x\|=\|y\|=1}|\langle A^*Ax,y\rangle|
=\sup_{\|x\|=\|y\|=1}|\langle Ax,Ay\rangle|\le \|A\|^2
$$
A: You seem to use use $J_x$ interchangeably with $J_X$ (note the potential for ambiguity as you also use lowercase $x$ to denote individual elements of the space $X$).  I will use the upper case letter when talking about $J_X$.
Continuing your chain of inequalities,
$$
\|A' J_X x\| \leq \|A'\| \|J_X x\| \leq \|A'\| \|x\|,
$$
where the first inequality follows from the definition of the operator norm, and the second uses the assumed fact that $J_X$ is an isometry.  So from this (and your chain of inequalities that preceded it) you can deduce $\|A^*\| \leq \|A'\|$ as desired, which fits into the rest of your argument the way that you want it to.
If it isn't clear, when I say that the first inequality in my above chain "follows from the definition of the operator norm," I am using the fact that $\|Ly\| \leq \|L\| \|y\|$ holds for for any Hilbert (or even Banach, or even normed linear) spaces $E$ and $F$, and any bounded operator $L: E \to F$, and any $y \in E$.  This is an immediate consequence of the definition of $\|L\|$ as an operator norm, using arguments simpler than the ones made up above.  And it is this fact that is being used up above, in the case $E = X'$, $F = X$, $L = A'$, and $y = J_X x$.
Remark: I wonder if some of this would be clearer if one were to notationally distinguish the norms of the space $X$, the space $X'$, and the space of linear operators from $X$ to $X'$ (endowed with the operator norm that it inherits from the norms of $X$ and $X'$).  Most textbooks don't seem to do this - as things do get notationally complex rather quickly - but without it, perhaps it is difficult to "see" in the notation what is actually going on.  Note for example in $\|A' J_X x\| \leq \|A'\| \|J_X x\|$ the notation $\|\cdot\|$ means something different each time it appears: first it denotes the norm on the space $X$, then it denotes the operator norm on the space of bounded linear operators from $X'$ to $X$, and then it denotes the norm on the space $X'$.
