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 ?