An upper bound for the number of divisors I've come across this problem in Murty & Carmen, exercise 1.5.3: show that there is a constant $c$ such that
$d(n)=O(\exp(\frac{c\log n}{\log\log n})))$ where $d(n)$ is the number of divisors of $n$
I've gotten most of the way through and have also looked through this article https://terrytao.wordpress.com/2008/09/23/the-divisor-bound/, but I don't understand this one step in their explanation:
$(6)\le O(1/\varepsilon)^{\exp(1/\varepsilon)} = \exp\exp(O(1/\varepsilon))$
Could anyone provide some justification as to why this would be true? I've tried playing around with the equality, but I've been having no luck. Thanks in advance!
 A: By unique factorization theorem, we can write $n$ into
$$
n=p_1^{r_1}p_2^{r_2}\cdots p_k^{r_k}q_1^{s_1}q_2^{s_2}\cdots q_m^{s_m}
$$
where $p_i$ denotes some prime $\le t$, and $q_j$ denote some prime $>t$. Thus, we have
$$
d(n)=\prod_i(r_i+1)\prod_j(s_j+1)
$$
By definition, we know that $k\le t$ and $r_i<\log_2 n$, so we have
\begin{aligned}
d(n)
&\le\left(1+\log_2n\right)^k\prod_j(s_j+1)\le(1+\log_2n)^t\prod_j2^{s_j} \\
&=(1+\log_2n)^t\prod_j q_j^{s_j\log2/\log q_j}<(1+\log_2n)^t\prod_jq_j^{s_j\log2/\log t} \\
&\le(1+\log_2n)^tn^{\log2/\log t}=\exp\{t\log(1+\log_2n)+\log2\log n/\log t\} \\
&\le2^{\log n/\log t+O(t\log\log n)}
\end{aligned}
Finally, setting $t=\log n/(\log\log n)^3$ gives $d(n)<2^{(1+\varepsilon)\log n/\log\log n}$ for large $n$.
For OP's question on the specific inequality, let $A_1,A_2,\dots$ denote the O constants then
\begin{aligned}
(A_1\varepsilon^{-1})^{\exp(\varepsilon^{-1})}
&=\exp[\exp(\varepsilon^{-1})\log(A_1\varepsilon^{-1})] \\
&<\exp[\exp(\varepsilon^{-1})\exp(A_2\varepsilon^{-1})] \\
&=\exp\{\exp[(1+A_2)\varepsilon^{-1}]\}=\exp\exp O(\varepsilon^{-1})
\end{aligned}
