Proof that the sequence $s_n=\ln(\ln(n))$ is unbounded. I am trying to construct a formal proof that
$$ \lim_{n\rightarrow\infty}s_n=+\infty $$
where $s_n=\ln(\ln(n))$. To do this I was going to show that the sequence is both monotonically increasing and unbounded. I was able to show it was increasing by writing the sequence as a composition of the increasing functions $f(x)=\ln(x)$ and $g(x)=\ln(x)$. Thus, the composition of two increasing functions will always be increasing. 
To show the sequence is unbounded, however, I am not quite sure where to begin. Intuitively I can see that this is true but I'm not quite sure how to begin the formal proof
 A: To show a sequence $f(n)$ is unbounded, we wish to show that for each $M$ there exists some $N$ such that there is some $n\geq N$ such that we have $|f(n)|>M$
Indeed, by taking $N=\lceil e^{e^{M}}\rceil$ and letting $n\geq N$ we have $\ln(\ln(n))\geq \ln(\ln(N))\geq \ln(\ln(e^{e^M}))=M$, proving the claim.
Depending on rigor, you may want to stop and prove each step in the inequality individually, but it follows as you say from the fact that $\ln$ is a monotonic increasing function, as is $\text{exp}$.
A: $\forall A>0$ you have to find $n_0$ such that when $n\ge n_0$, $f(n)\ge A$. You can choose $n_0=\lfloor \exp(\exp A)\rfloor +1$
A: Let $M \in \mathbb{R}$. Choose an $N \in \mathbb{N}$ such that $N > e^{e^{M}}$ ($N$ exists by the Archimedean Property). Let $n \in \mathbb{N}$, and suppose $n \geq N$. Then $$s_n = \ln (\ln n) \geq \ln (\ln N) > \ln (\ln e^{e^M}) = \ln e^M = M.$$ Thus, $\lim s_n = + \infty$.
A: For unbounded $m$, $\log(m)$ is unbounded. Hence with $m:=\log(n)$, for unbounded $\log(n)$, $\log(\log(n))$ is unbounded. Hence for unbounded $n$, $\log(\log(n))$ is unbounded.
