Prove that for every $ϵ>0$ there exist infinitely many natural numbers $n$, such that $τ(n)\geq 2^{(1-ϵ)(\log(n)/\log(\log(n))}$. Prove that for every $ϵ>0$ there exist infinitely many natural numbers $n$, such that $$\tau(n)\geq 2^{(1-ϵ)\frac{\log(n)}{\log\log(n)}}$$
It is obvious to look at squarefree naturals of the form $n=p_1 p_2\cdots p_k$ where $p_1,p_2,\ldots,p_k$ are first $k$ prime numbers. Such naturals have $\tau(n)=2^k$.
So I have to prove that for k large enough, $k>(1-ϵ)\frac{\log(n)}{\log(\log(n))}$.
I thought about putting exp on both sides so we have to prove 
$\exp(k)>n^{\frac{1-ϵ}{\log(\log(n))}}$.
There must be a chain of inequalities starting from $n^{\frac{1-ϵ}{\log(\log(n))}}<\cdots<\cdots<\cdots<\exp(k)$. I was trying to do it yesterday but my inequalities were tnot proper. Of course I can write "$1$" instead of "$1-ϵ$" or write $\log(\log(2))$ or $\log(\log(p_k))$ instead of $\log(\log(n))$ in denominator but that directs me to nothing. Later all I can do with $n$ is saying that $n<p_k^k$ but this is too strong - because this side will be bigger than $\exp(k)$.
Any hints?
 A: You could use Wigert's result
$$\limsup_{n\to\infty}\frac{\log \tau(n)}{\log n/\log\log n}=\log2 \Rightarrow \\
\limsup_{n\to\infty}\frac{\log \tau(n)}{\log2 \cdot \log n/\log\log n}=1 \Rightarrow \\
\exists n \text{ s.t. } \frac{\log \tau(n)}{\log2 \cdot \log n/\log\log n}> 1-\varepsilon \Rightarrow \\
\tau(n)>2^{(1-\varepsilon)\frac{\log{n}}{\log{\log{n}}}}$$

Or use Chebyshev function and the fact that $\forall \varepsilon>0$, $\exists x(\varepsilon)>0$ s.t. $\forall x > x(\varepsilon)$
$$e^{(1-\varepsilon)x}<\prod\limits_{p\leq x}p< e^{(1+\varepsilon)x} \Rightarrow $$
$$(1-\varepsilon)x<\log{\prod\limits_{p\leq x}p}< (1+\varepsilon)x \Rightarrow 
\color{blue}{x > \frac{\log{\prod\limits_{p\leq x}p}}{1+\varepsilon}} \tag{1}$$
Use PNT or better $$\color{blue}{\pi(x)>\frac{x}{\log{x}}} \tag{2}$$
Note that $\frac{x}{\log{x}}$ is ascending for large enough $x$'s:
$$\frac{x}{\log{x}}\overset{\color{red}{(1)}}{>} \frac{\frac{\log{\prod\limits_{p\leq x}p}}{1+\varepsilon}}{\log{\frac{\log{\prod\limits_{p\leq x}p}}{1+\varepsilon}}}\tag{3}$$
Then, by noting $n_x=\prod\limits_{p\leq x}p$ and considering that $2^x$ is ascending:
$$\tau(n_x)=\tau\left(\prod\limits_{p\leq x}p\right)= 
2^{\pi(x)}\overset{\color{red}{(2)}}{>}
2^{\frac{x}{\log{x}}} \overset{\color{red}{(3)}}{>} \\
2^{\frac{1}{1+\varepsilon}\frac{\log{\prod\limits_{p\leq x}p}}{\log{\left(\log{\prod\limits_{p\leq x}p}\right)-\log{(1+\varepsilon)}}}}=
2^{\frac{1}{1+\varepsilon}\frac{\log{n_x}}{\log{\log{n_x}}-\log{(1+\varepsilon)}}}> 
2^{(1-\varepsilon)\frac{\log{n_x}}{\log{\log{n_x}}}}$$
because $1>1-\varepsilon^2=(1+\varepsilon)(1-\varepsilon) \Rightarrow \frac{1}{1+\varepsilon}>1-\varepsilon$. We can build an infinite sequence of such $n_x$'s.
