Prove cardinals equality

Question

Let $x,c$ two cardinals, such that:
$1\lt x \le c$
$c^2 = c$

Prove:
$x^c = 2^c$

So, from the second statement, we know $c$ is infinite, because it cannot be true for a finite cardinal.

I know the "tools" to prove it must be:

1. CBS theorem
2. cardinals arithmatic
3. squeezing theorem

I tried playing with these, but all I got is bunch of inequalities didn't lead to the answer.

EDIT: What I did so far:
$$\begin{array}{l}x \le b\\{2^x} \le {2^b}\\{\left( {{2^x}} \right)^b} \le {\left( {{2^b}} \right)^b}\\{2^{xb}} \le {2^{bb}}\end{array}$$

I'm pretty much stuck at this point.

Answer 1

HINT: Remember the laws of exponentiation:

1. $a\leq b\implies 2^a\leq 2^b$.
2. $(a^b)^c=a^{bc}$.

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To the edit:

HINT: Show that if $x\leq c=c^2\implies xc=c$.

Answer 2

It’s easy to find an injection from ${}^c2$ to ${}^cx$; the hard part is finding one from ${}^cx$ to ${}^c2$. You also have an easy injection from ${}^cx$ to ${}^cc$, so all you really need is an injection from ${}^cc$ to ${}^c2$:

$${}^cc\to{}^c({}^c2)\to{}^{c\times c}2\to{}^c2$$