Equality of power towers : $a\uparrow\uparrow m=b\uparrow \uparrow n$

Question

Suppose, $a,m,b,n$ are natural numbers greater than $1$.

If we have $$a\uparrow\uparrow m=b\uparrow\uparrow n$$

can we conclude $a=b$ and $m=n$ ?

$a\uparrow \uparrow m$ is a powertower of $m$ $a's$ and $b\uparrow\uparrow n$ a power tower of $n$ $b's$.

I proved $b^b\ne a^{a^a}$ for natural numbers $a,b>1$.

This is my proof :

Set $s:=\frac{ln(b)}{ln(a)}$. So, we have $a^s=b$. Therefore, we have

$$a^{a^a}=(a^s)^{(a^s)}=a^{sa^s}$$ , so $a^a=sa^s=sb$, we can conclude that $s$ is rational.

If $s\ge a$, then $sa^s\ge aa^a>a^a$. If $s\le a-1$, then $sa^s=(a-1)a^{a-1}<aa^{a-1}=a^a$

Hence, we can conclude $a-1<s<a$ implying that $s$ is not an integer.

Finally, assume $s=\frac{m}{n}$ with coprime $m$ and $n$. Then, we have

$$(\frac{m}{n})^na^{s\ n}=a^{an}$$, implying $$(\frac{m}{n})^n=a^{an-m}$$

Because of $a>s$ we have $an-m>sn-m=0$. Therefore, the left side of the last equation is not an integer, while the right side is an integer. This contradiction finally proves the claim.

Questions:

Is this proof correct ?

Is there an easier proof that $a^{a^a}=b^b$ cannot hold for natural numbers $a,b>1$ ? (For example, using the $p$-adic value of $a^{a^a}$ and $b^b$) ?

Is there a proof that we can extend to show the more general initial claim ?

Answer 1

One can prove your initial claim by proving the following more general statement:

Theorem. $a^{a^b} = c^{c^{c^d}}$ with $a > c \ge 2$ and $b,d \ge 1$ has no solutions.

Proof. Write $a^z = c^y$ with $\text{gcd}(z,y) = 1$. Then $a^{\frac{1}{y}} = c^{\frac{1}{z}} = x$ is an integer, and $a = x^y, c = x^z$. So

$$x^{y x^{yb}} = x^{z x^{z x^{zd}}}$$

$$y x ^{yb} = z x^{z x^{zd}}$$

hence $\frac{y}{z}$ is a positive power of $x$; let $y = x^w$. Then

$$z x^w x^{zx^w b} = zx^{zx^{zd}}$$ $$w + zx^w b = zx^{zd}$$

But, since $w < zx^w$, $w + zx^w b$ lies between $zx^w b$ and $zx^w(b+1)$, and therefore cannot be of the form $zx^{zd}$.