Let $p_{i}$ be the $i^{th}$ prime. Let $M_{k}$ be the number of $5^{th}$ power free numbers in the set $\{1, 2, \ldots, (p_{1}^{4}p_{2}^{4}\cdots p_{k}^{4}) \}$, that are divisible by some prime $p \not\in\{p_{1}, p_{2}, \ldots, p_{k}\} $. I am interested in finding $\lim_{k\rightarrow\infty}\frac{M_{k}}{p_{1}^{4}p_{2}^{4}\cdots p_{k}^{4}}$.
I tried using prime number theorem but could not get a useful result. Any help would be greatly appreciated.
For any positive integer $q$, we know that the number of integers $n\in\{1,2,\dots,q\}$ that are relatively prime to $q$ equals $\phi(q)$. Since being relatively prime to $q$ depends only on the residue class of $n$ (mod $q$), it follows that for any multiple $Q$ of $q$, the number of integers $n\in\{1,2,\dots,Q\}$ that are relatively prime to $q$ equals $Q\phi(q)/q$. Taking $q=p_1\cdots p_k$ and $Q=q^4$, we see that the proportion of integers $n\in\{1,2,\dots,p_1^4\cdots p_k^4\}$ not divisible by any of $p_1,\dots,p_k$ is exactly $1-\frac{\phi(p_1\cdots p_k)}{p_1\cdots p_k}$, which tends to $0$ as $k$ tends to infinity.
Consequently, $M_k$ is the same as the proportion of integers not divisible by any of $p_{k+1}^5,p_{k+2}^5,\dots$, which one can show approaches $1$ as $k\to\infty$. Indeed, a more careful analysis shows that $$ M_k = 1 - \frac{\phi(p_1\cdots p_k)}{p_1\cdots p_k} + O(p_{k+1}^{-4}) \sim 1 - \frac1{e^\gamma\log k}, $$ where $\gamma$ is Euler's constant.