Letting $L = \limsup_n \frac{X_n - \log n}{\log \log n}$, for $\{X_n\}_{n\geq 1}$ iid exponentially distributed random variables with rate $1$, I think I am able to show that $L=1$ (almost everywhere). Assuming I hadn't made some mistake,it boiled down to a direct application of the Borel-Cantelli lemmas via events $E_n = \left\{\frac{X_n - \log n}{\log \log n} > a\right\}$, $$ \sum P\left(E_n\right) = \sum P \left(\frac{X_n - \log n}{\log \log n} > a\right) = \sum \frac{1}{n (\log n)^a} = \begin{cases} <\infty & \text{when $a > 1$} \\ = \infty & \text{when $a \leq 1$} \end{cases} $$ Therefore, by BC $$ \implies P(E_n, \text{i.o}) \equiv P(\limsup_n E_n) = \begin{cases} 0 & a > 1 \\ 1 & a \leq 1 \end{cases} $$ So, as $a \to 1$, $$ 1 = P\left( \limsup_n \left\{ \frac{X_n - \log n}{\log \log n} > 1 \right\} \right) \leq P\left( \limsup_n \left\{\frac{X_n - \log n}{\log \log n} \right\} \geq 1 \right) = P(L \geq 1) $$ Similarly, we can end up showing that $P(L > 1) = 0$, giving us the desired result.
My question is a matter of intuition. I (think) I understand the mathematics, but I just cannot wrap my head around how this is at all possible at any level beyond the symbolic manipulation. If I think about sampling/generating exponential deviates $X_n \stackrel{\text{iid}}{\sim} \text{Exp}(1)$, how is it possible that the ratio $\frac{X_n(\omega) - \log n}{\log \log n}$ could possibly approach 1 (for all $\omega$ except those on measure zero sets). I've tried simulating exponential samples and calculating this ratio, keeping $\omega$ fixed (i.e. the PRNG seed fixed) across $n = 1, ..., \texttt{nmax}$. Regardless of how many samples I do, I always end up seeing the behaviour one would expect from, say $\frac{x - \log n}{\log \log n}$ as $n\to\infty$.
Clearly something is either wrong in my calculations above, or something is wrong with my intuition. Any insight would be appreciated!
You ask "how is it possible that the ratio $r_n:=\frac{X_n(\omega) - \log n}{\log \log n}$ could possibly approach 1 (for all $\omega$ except those on measure zero sets)?" but that is not what limsup means. What is true, and follows from your calculation, is that for every fixed $m$, the sequence $$A_m(N):= \max_{n \in [m,N]} \frac{X_n(\omega) - \log n}{\log \log n}$$ approaches 1 as $N \to \infty$ (for all $\omega$ except those on a set of measure zero). In other words, almost surely, there is a subsequence of $r_n$ that tends to 1, and no subsequence can tend to a higher limit.