Does $(\ln(a))^b$ ($a\in\mathbb Z_{>1}$; $b\in\mathbb Z_{\ne0}$) have a unique representation in this form? I just want to deal numbers in a form of $(\ln(a))^b$ ($a\in\mathbb Z_{>1}$; $b\in\mathbb Z_{\ne0}$), but I wondered if each number in this form has a unique form.
In other words: is a function $f$ defined by $f(a,b)=(\ln(a))^b$ ($a\in\mathbb Z_{>1}$; $b\in\mathbb Z_{\ne0}$) injective?
I found that this question is equivalent to: is $\frac{\ln(\ln(p))}{\ln(\ln(q))}$($p,q\in\mathbb Z_{>1}$; $p\ne q$) always irrational?
But I have no idea how to go further.
 A: My comments
I like the question.  To recap:
$$
(\ln p)^b = (\ln q)^d,\qquad p,q \in \mathbb Z_{> 1}, \quad p \ne q, \quad b,d \in \mathbb Z_{\ne 0}\\
\Longleftrightarrow\qquad
\ln p = (\ln q)^r,\qquad p,q \in \mathbb Z_{> 1}, \quad p \ne q, \quad r \in \mathbb Q
\\
\Longleftrightarrow\qquad
\frac{\ln\ln p}{\ln\ln q} = r,\qquad p,q \in \mathbb Z_{> 1}, \quad p \ne q, \quad r \in \mathbb Q
$$
Some cases I can do.
Case 1. $s \in \mathbb Z_{> 1}$ and $p = s^\alpha, q=s^\beta$ for $\alpha,\beta \in \mathbb Z_{> 0}$  [I hinted $p=2, q=4$ above.]  Then
$$
\frac{\ln\ln p}{\ln\ln q} \in \mathbb Q
\\
\Longleftrightarrow\qquad
\frac{\ln \alpha +\ln\ln s}{\ln\beta + \ln\ln s} \in \mathbb Q
\\
\Longleftrightarrow\qquad
\frac{\ln \alpha +\ln\ln s}{\ln\beta + \ln\ln s} \in \mathbb Q
\\
\Longleftrightarrow\qquad
\frac{\ln \alpha -\ln \beta}{\ln\beta + \ln\ln s} \in \mathbb Q
\\
\Longleftrightarrow\qquad
\frac{\ln\beta + \ln\ln s}{\ln \alpha -\ln \beta} \in \mathbb Q
\\
\Longleftrightarrow\qquad
\frac{\ln\beta + \ln\ln s}{\ln(\alpha/\beta)} \in \mathbb Q
$$
So write $\gamma = (\ln\beta + \ln\ln s)/(\ln(\alpha/\beta))$.  Assume $\gamma$ is rational.  Note
$$
\left(\frac{\alpha}{\beta}\right)^\gamma = \exp\big(\gamma\ln(\alpha/\beta)\big) =
\exp\big(\ln\beta + \ln\ln s\big) =
\beta \ln s
$$
But note: $\alpha/\beta \in \mathbb Q, \gamma \in \mathbb Q$ so
$(\alpha/\beta)^\gamma$ is algebraic.  On the other hand, $s \in \mathbb Z_{> 1}$ so $\ln s$ is transcendental [by Lindemann-Weierstrass ]; also $\beta \in \mathbb Z_{> 0}$, so $\beta\ln s$ is transcendental.  A contradiction.

What if $p$ and $q$ are not both powers of the same integer $s$?  I do not know a proof in that case.  But it does follow from Schanuel's conjecture .
Suppose
$$
\ln p = (\ln q)^r,\qquad p,q \in \mathbb Z_{> 1}, \quad p \ne q, \quad r \in \mathbb Q
\tag{$*$}
$$
but $p,q$ are not both powers of the same $s \in \mathbb Z_{> 1}$.
Now $\ln p$ and $\ln q$ are linearly independent over $\mathbb Q$ by unique factorization.  Schanuel's conjecture then asserts that the field
$$
\mathbb Q\big(\ln p, \ln q, p, q\big)
$$
has transcencence degree at least $2$.  However, $p, q \in \mathbb Q$ and $\ln p, \ln q$ are algebraically related by $(*)$, so we have a contradiction.
