Tetration convergence: prove $\lim_{x\rightarrow0} {}^{n}x = \begin{cases} 1, & n \text{ even} \\ 0, & n \text{ odd} \end{cases}$

As mentioned in the comments, you want to approach this by induction. What is missing in to show that as $x\to 0^+, 1^-$, then $x^x$ approaches $1$ from below.

Step 1: $$\begin{align}\lim_{x\to 0}x^x&=\lim_{x\to 0}e^{x\ln x}\\&=e^{\lim_{x\to 0}x\ln x}\\&=\exp\left({\lim_{x\to 0}\frac{\ln x}{1/x}}\right)\\&=\exp\left(\lim_{x\to 0}\frac{1/x}{-1/x^2}\right)\\&=e^0\\&=1 \end{align}$$

Step 2: $$\begin{align}\frac d{dx}x^x&=\frac d{dx}e^{x\ln x}\\&=e^{x\ln x}\frac d{dx}(x \ln x)\\&=x^x(\ln x+1) \end{align}$$

Step 3:

Using step 1 above, the limit of the derivative of $x^x$ when $x\to 0^+$ is $-\infty$. But the only thing we care is that it is negative

Step 4:

If $1/e<x$ then the derivative of $x^x$ will be positive. At $x=1$, $x^x=1$, so for $x$ slightly less than $1$, $x^x$ is increasing towards $1$.

Note that:

$$x^x=\exp(x\ln(x))=1+\mathcal O(x\ln(x))\tag{$x\to0$}$$

and in general,

\begin{align}x^{1+\mathcal O(x\ln(x))}&=x\cdot x^{\mathcal O(x\ln(x))}\\&=x\exp(\mathcal O(x\ln^2(x)))\\&\sim x\tag{$x\to0$}\end{align}

and

\begin{align}x^{\mathcal O(x)}&=\exp(\mathcal O(x\ln(x)))\\&=1+\mathcal O(x\ln(x))\tag{$x\to0$}\end{align}

from which you can easily see that it will alternate between being approximately $x$ and $1+\mathcal O(x\ln(x))$, and hence the limit alternates between $0$ and $1$.