# How to prove that $e = \lim_{n \to \infty} (\sqrt[n]{n})^{\pi(n)} = \lim_{n \to \infty} \sqrt[n]{n\#}$?

While reading this post, I stumbled across these definitions (Wiki_german)

$$e = \lim_{n \to \infty} \sqrt[n]{n\#}$$

and

$$e = \lim_{n \to \infty} (\sqrt[n]{n})^{\pi(n)}.$$

The last one seems interesting, since $\lim_{n \to \infty} (\sqrt[n]{n})=1$, proven here.

How to prove these?

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In the second one, you can use that $\pi(n)\sim n/\log(n)$ asymptotically. – Raskolnikov Jan 5 '12 at 10:11
Regarding your comment, note that $\pi(n)\to\infty$, so in order for $\sqrt[n]{n}^{\pi(n)}$ to have a finite limit it is necessary for $\sqrt[n]{n}$ to converge to $1$. – Jonas Meyer Jan 5 '12 at 10:13
@Raskolnikov: So infact one doesn't really need $\pi(n)$ to prove it? – draks ... Jan 5 '12 at 13:39
Let's just say that it is an awkward way to rephrase a property of the prime-counting function. Same is probably true for the primoral $n\#$. – Raskolnikov Jan 5 '12 at 13:43
Definitions? Of what are those two formulas "definitions"? Anyway, the product of the primes up to $n$ is asymptotic to $e^n$ - I believe this is at the same level of difficulty as the Prime Number Theorem - that should get you the first equation. – Gerry Myerson Mar 26 '12 at 12:10