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Show that every large integer has a large prime-power factor

That is, if $P(n)$ designates the largest number $p^a$ which divides $n$, then $\lim_{n\to\infty}P(n)=\infty.$

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HINT Show that for every $M \in \mathbb N$, there exists $N = N(M) \in \mathbb N$ such that $ P(n) \leq M \implies n \leq N .$ Why does this statement prove that $P(n) \to \infty$ as $n \to \infty$?

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    @Z_Fang Well, actually, there's not much to prove there. The first sentence is essentially the definition of "$P(n) \to \infty$ as $n \to \infty$".2011-10-24
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$\rm\bf Guide$: Let $p^a$ be a prime power. There are only finitely many prime powers less than it, and every natural can be uniquely written as a product of prime powers. Use this to show that the number of naturals $n\in\mathbb{N}$ such that $P(n)=p^a$ is finite, no matter what $p^a$ is. Along these lines, for given $p^a$, what explicitly is the largest $n$ such that $P(n)=p^a$? What does this mean for $P(\cdot)$ for all $\cdot>n$? How does this show that $\lim\limits_{n\to\infty} P(n)=\infty$?