Numbers divisible by the square of their largest prime factor Let $p(n)$ be greatest prime factor of $n$, denote $A=\{n\mid p^2(n)\mid n,n\in \mathbb N\}.$
$A=\{4,8,9,16,18,25,27,32,36,49,50,\cdots\},$ see also A070003.
Define $f(x)=\sum_{\substack{n\leq x\\n\in A}}1.$ Erdős proved that
$$f(x)=x \cdot e^{-(1 + o(1))\sqrt{\log x \log \log x}}.\tag1$$
Hence $f(x)=o(x),$
$$\lim_{x\to \infty}\frac{f(x)}x=0,\tag2$$
since $(2)$ is  easier than $(1)$, can you prove $(2)$ without use $(1)$?

If $n\in A$ then $n$ is divisible by $p(n)^2$, so that $n$ cannot divide $P(n)!$.
Hence it is enough to show that $S(x)=\{n \le x \mid n \nmid p(n)!\}$ satisfies $|S(x)|=o(x)$ for $x\to \infty$. This was shown by I. Kastanas, The smallest factorial that is a multiple of $n$ (solution to problem 6674), Amer. Math. Monthly 101 (1994), p.179. On the other hand, already some of the arguments Erdős gives for (1) are already enough to conclude (2), but since the question says "without use (1)", the other way seems better.
