Out of boredom, I decided to recall the following equation:

$$x^{x^{x\cdots}} = 2.$$

Which, I simply rewrote like this:

$x^2 = 2$, and therefore $x = \sqrt{2}$. Then I took a look at the more general form:

$$x^{x^{x\cdots}} = p.$$

I then concluded that $x = (p)^{\frac{1}{p}}$, and this was the solution set for all $p$. Then I thought for a while and determined that there are multiple values where this function equals the square root of $2$, so its not injective. However, this led to an awkward statemend based on the fact that $\sqrt{2} = (4)^{\frac{1}{4}}$. We deduce the following:

The equation $x^{x^{x\cdots}} = 2$ and the equation $x^{x^{x\cdots}} = 4$ have the same solution, namely $x = \sqrt{2}$. So we conclude that $2 = 4$.

I showed this to friend of mine, and he conjectured that there are no valid solutions for all $p > e$. To prove this, I tried writing the following:

I tried differentiating $f(x) = (x)^{(1/x)}$ and setting it equal to $0$ to find the maximum. I determined that there was a maximum at the point where $x = e$. Does this mean that the maximum possible solution $x$ for the infinite power is $e^{(1/e)}$ therefore showing that the maximum value for $p$ is indeed $p = e$?