Proving $d(m) \ll m^{\epsilon}$ I am reading a text currently, and in the proof of lemma 3.1 it requires us to show that for any $\epsilon > 0$, we have $d(m) \ll m^{\epsilon}$ where $d(m)$ is the number of divisors. The text then writes

We suppose that $m = p_1^{\lambda_1} p_2^{\lambda_2} \cdots$, and note that
$$\frac{d(m)}{m^{\epsilon}} = \prod_i \frac{\lambda_i + 1}{p_i^{\epsilon \lambda_i}} \leq \prod_{p_i \leq 2^{1 / \epsilon}} \frac{\lambda_i + 1}{2^{\epsilon \lambda_i}} \leq C(\epsilon)
$$
Since $2^{-\epsilon\lambda}(\lambda + 1)$ is bounded above for $\lambda > 0$.

I don't understand the step where he changed from taking the product over all prime powers to only those with $p_i \leq 2^{1 / \epsilon}$. I had the idea that it might be because he is taking an upper bound by only considering the terms $\geq 1$, and hence we want
$$
\frac{\lambda_i + 1}{p_i^{\epsilon\lambda_i}} \geq 1
$$
However, this doesn't seem to give anything related to $p_i \leq 2^{1 / \epsilon}$. Can someone help figure what the author meant?
Thank you!
Gareth
 A: The fact is that there is only a finite amount of primes which are smaller than $2^{1/ \varepsilon}$. Hence, for all $\varepsilon >0$ you can consider
$$M= \max_{\lambda \ge 0} 2^{- \varepsilon \lambda}(\lambda +1)$$
$$N= |\{ \mbox{primes smaller than } 2^{1/ \varepsilon}\}|$$
Then
$$\prod_{p_i \le 2^{1/ \varepsilon}} \frac{\lambda_i+1}{p_i^{\varepsilon \lambda_i}} \le \prod_{p_i \le 2^{1/ \varepsilon}} \frac{\lambda_i+1}{2^{\varepsilon \lambda_i}} \le \prod_{p_i \le 2^{1/ \varepsilon}} M = M^N$$
$M^N$ is a constant depending only on $\varepsilon$ and not on $m$.
On the other hand you should notice that for all $\lambda \ge 0$
$$\lambda +1 \le 2^{\lambda}$$
Hence
$$\prod_{p_i > 2^{1/ \varepsilon}}\frac{\lambda_i+1}{p_i^{\varepsilon \lambda_i}} \le \prod_{p_i > 2^{1/ \varepsilon}}\frac{\lambda_i +1}{2^{\lambda_1}} \le \prod_{p_i > 2^{1/ \varepsilon}} 1 =1 $$
Putting everything together you get
$$\frac{d(m)}{m^{\varepsilon}} \le M^N \cdot 1 =M^N$$
A: Nevermind, you simply solve the inequality...
$$
p_i^{\epsilon} \leq \sqrt[\lambda_i]{\lambda_i + 1}
$$
And the right hand side is bounded by $2$. Even though we include more terms than just the ones with $\frac{\lambda_i + 1}{p_i^{\epsilon \lambda_i}} \geq 1$, it's still an upper bound since we included all of them (and multiplied by more terms that was already in the original product). Oops :P
