I am guessing that this is the exact same solution the OP (TheBigOne) got, but I am putting my answer here, so that the thread is answered. Even though this proof requires some non-elementary knowledge (Bertrand's Postulate), it is still a proof. (Although I disagree that Bertrand's Postulate is not elementary. The proof I know is quite elementary, and as i707107 said, Bertrand's Postulate is very widely known.) The solution also utilizes the fact that there are infinitely many prime natural numbers congruent to $2$ modulo $3$. (This fact has an elementary proof.)
First, let $L_k:=\text{lcm}(1,2,\ldots,k)$ for $k=1,2,3,\ldots$. Note that $L_1=1$ and $L_k\geq k(k-1)$ for $k\geq 2$. That is, for each integer $N>0$, $$\sum_{k=1}^N\,\frac{1}{L_k}\leq 1+\sum_{k=2}^N\,\frac{1}{k(k-1)}\leq 1+\sum_{k=2}^\infty\,\frac{1}{k(k-1)}=2\,.$$
This means $S:=\sum\limits_{k=1}^\infty\,\dfrac{1}{L_k}$ exists and is a positive real number less than $2$. (WolframAlpha states that $S\approx 1.77111$.)
We argue by contradiction. Suppose contrary that $S=\dfrac{a}{b}$ for some relatively prime positive integers $a$ and $b$. Let $p_1,p_2,p_3,\ldots$ be the increasing sequence of all prime natural numbers greater than $b$. Using Bertrand's Postulate, $$p_r<p_{r+1}<2\,p_r$$ for all $r=1,2,3,\ldots$. Thus, $$p_{r+1}-p_r\leq p_r-1$$ for each positive integer $r$. Note that, for infinitely many positive integers $r$, it holds that $p_r\equiv 2\pmod{3}$. Therefore, the equality $p_{r+1}-p_r=p_r-1$ does not happen (or else, $p_{r+1}\equiv 0\pmod{3}$). Hence, $p_{r+1}-p_r<p_r-1$ for infinitely many such $r$.
Now,
$$\begin{align}
L_{p_1-1}\,\left(S-\sum_{k=1}^{p_1-1}\,\frac{1}{L_k}\right)&\leq \sum_{r=1}^\infty\,\sum_{k=p_r}^{p_{r+1}-1}\,\frac{L_{p_1-1}}{L_k}\leq \sum_{r=1}^\infty\,\sum_{k=p_r}^{p_{r+1}-1}\,\frac{1}{p_1p_2\cdots p_r}
\\
&=\sum_{r=1}^\infty\,\frac{p_{r+1}-p_r}{p_1p_2\cdots p_r}<\sum_{r=1}^\infty\,\frac{p_{r}-1}{p_1p_2\ldots p_r}
\\
&=\left(1-\frac{1}{p_1}\right)+\left(\frac{1}{p_1}-\frac{1}{p_1p_2}\right)+\left(\frac{1}{p_1p_2}-\frac{1}{p_1p_2p_3}\right)+\ldots
\\
&=1\,.
\end{align}$$
This is a contradiction, as $b\mid L_{p-1}$ and $S>\sum\limits_{k=1}^{p_1-1}\,\frac{1}{L_k}$, which means $L_{p-1}\,\left(S-\sum\limits_{k=1}^{p_1-1}\,\frac{1}{L_k}\right)$ is a positive integer. Therefore, $S$ cannot be a rational number.
Let $x = \frac{a}{b}$ where $a,b$ are coprime. Then we have $(\frac{a}{b})^{\frac{a}{b}} = n$. By lifting both sides to the power of $b$ we have:
$$\left(\frac{a}{b}\right)^a = n^b \\ \implies a^a = n^b \cdot b^a $$
Assume $b\neq 1$. Then there exists a prime $p$ such that $p$ divides $b$. Therefore $p^a$ divides $b^a$, thus $p$ divides $a^a$. But if $a$ is not divisible by $p$, then neither is $a^a$. Therefore $a$ is divisible by $p$. As both $a$ and $b$ are divisible by $p$, they cannot be coprime. Contradiction.
Therefore $b = 1$. Thus integral solutions are the only rational solutions to $x^x = n$ .
Best Answer
We can read in the Wikipedia article
Here the phrase unexpected nature of the result addresses the unexpected technique to tackle the problem using methods known from showing the irrationality of $\zeta(2)$, while the experts rather expected methods similar to those which were used in trying to show that the quantities \begin{align*} \frac{\zeta(2n+1)}{\pi^{2n+1}} \end{align*} are transcendental.
At the end of this paragraph there is also a nice reference to the paper A proof that Euler missed... by A. v. d. Poorten. The author tells us:
Conclusion: What we can read here addresses the interesting techniques to prove the irrationality of $\zeta(3)$. Irrationality was not seriously questioned.