Proving $x*\ln(\ln(x))>\ln(\ln(x\#))$ for $\infty>$x>e$

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Let $x \in \mathbb{N}$.

For $\infty>x>e$,

Let $x$ be the largest prime in the factorization of the primorial $x\#$.

How does one go about proving $x*\ln(\ln(x))>\ln(\ln(x\#))$?

The plot of LHS$-$RHS looks like it will be always positive, so heuristically it seems that there should be a simple proof, but so far it has eluded me. Here is my attempt:

From the definition of the primorial, we easily have

$n*\ln(x)>\ln(x\#)$

where $n$ is the number of primes in $x$. Then,

$\ln(n)+\ln(\ln(x))>\ln(\ln(x\#))$

$x*\ln(\ln(x))>x*(\ln(\ln(x\#))-\ln(n))>\ln(\ln(x\#))-\ln(n)$

So, at least it's true for sufficiently small $n$.

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$\ln(\ln y)$ is unbounded, that is, it can be made as large as you like for a large enough $y$. Without any other constraint on $x$ and $y$, the problem seems incomplete