sequences and series - How to prove that $displaystyle lim_{n to infty} sqrt[n]{n}=1$?

I have seen this fact thrown around a lot and never really stopped to prove it; plugging in a few values convinced me of its truth. But I would like to see the result proved. To be clear, this is not homework, just human curiosity; I'm looking for any nice proof. Thanks.

Answer

Rewrite as

$$\sqrt[n]{n}=n^{1/n} = e^{\ln(n)/n}$$
Now, you can write that

$$\lim_{n\to\infty} \sqrt[n]{n} = \lim_{n\to\infty} e^{\ln(n)/n} = e^{\lim_{n\to\infty}\ln(n)/n}$$
Looking at the exponent, you have (using L'Hopital's Rule)
$$\lim_{n\to\infty}\frac{\ln(n)}n = \lim_{n\to\infty} \frac{1/n}{1}=\lim_{n\to\infty} \frac1n = 0$$
Therefore, you have
$$\lim_{n\to\infty} \sqrt[n]{n} = e^0 = 1$$