Optimal bounds for the product of the divisor function $d(n)$ in short intervals Let $d(n)$ denote the number of divisors of a positive integer $n$.
It is pretty obvious that $d(n) \ge 2$ for any given number $n \ge 2$, since every number is divisible by $1$ and itself. $2$ is also the best lower bound since there is an infinite amount of prime numbers. One can also quite easily see that $d(2n)d(2n+1) \ge 8$ for any number $n \ge 3$ since $2n$ is divisible by $1$, $2$, $n$ and $2n$, so $d(2n) \ge 4$, while $d(2n+1) \ge 2$. To my knowledge, it is not known whether that bound is optimal or not.
To generalize that result, let $s(m)$ denote the optimal bound for $\displaystyle\prod_{k=0}^{m-1} d(mn+k)$ for $n$ big enough, meaning $$s(m)=\liminf_{n \to \infty} \prod_{k=0}^{m-1} d(mn+k)$$
I'm now interested in the growth rate of $s(m)$, especially whether or not $\displaystyle \max_m(s(m)^\frac1m) = \infty$
It is clear that $s(m) \ge 2^m$, so it grows at least exponentially. Are there any better lower bounds?
Edit:
Does it make a difference whether one starts at $mn+0$ and ends at $mn+m-1$, or multiplies over any arithmetic progression of length $m$ for big enough $n$? If it does, how does the lower bound $s(m)$ change?
 A: The answer is yes, I'll show that the limit you are interested in does indeed tend to infinity as $m\rightarrow\infty$.
Define $$s_{m}(n)=\prod_{i=0}^{m-1}\left(d(n+i)\right).\ \ \ \ \ \ \ \ \ \ \ \ \ (1)$$
We have the following Theorem:

Theorem 1:  For constants in the big-$O$ term independent of $n$, as $m\rightarrow\infty$, $$s_{m}(n)\geq2^{m\log\log m+O(m)}.$$ In particular, uniformely in $n$, $$\lim_{m\rightarrow\infty}s_{m}(n)^{\frac{1}{m}}=\infty$$.

Since this Theorem holds uniformely in $n$, it answers the questions in the edit as well.
Proof of Theorem: For any prime $p$, at least $\lfloor\frac{m}{p}\rfloor$ of the integers $n,n+1,\dots,n+m-1$ will be divisible by $p$. It follows that for the prime $p$, at least $\lfloor\frac{m}{p}\rfloor$ of the integers $n,n+1,\dots,n+m-1$ will have $d(p)\geq2$. Since $$d(n)=\prod_{\begin{array}{c}
q|n\\
q\ \text{prime}
\end{array}}d(q)$$ for any $n$, we may change the order of the products and write $$s_{m}(n)=\prod_{i=0}^{m-1}\prod_{p|n+i}d(p)=\prod_{p}\prod_{\begin{array}{c}
i=0\\
p|n+i
\end{array}}^{m-1}d(p)$$ $$\geq\prod_{p}\prod_{\begin{array}{c}
i=0\\
p|n+i
\end{array}}^{m-1}2=\prod_{p}2^{\#\left\{ i:\ p|n+i\right\}}.$$ For each prime $p$, $\#\left\{ i:\ p|n+i\right\}\geq \lfloor\frac{m}{p}\rfloor$, and thus $$s_{m}(n)\geq\prod_{p}2^{\lfloor\frac{m}{p}\rfloor}=2^{\sum_{p\leq m}\lfloor\frac{m}{p}\rfloor}.$$ Mertens Second Theorem states that $$\sum_{p\leq m}\frac{1}{p}=\log \log m + M +o(1)$$ where $M$ is the Meissel–Mertens constant, and hence we have $$\sum_{p\leq m}\lfloor\frac{m}{p}\rfloor=m\log\log m+O(m),$$  and the result follows.
Remark: The sum $\sum_{p\leq m}\lfloor\frac{m}{p}\rfloor=\sum_{\leq x}\omega(n)$ where $\omega(n)$ is the distinct prime divisors function.
