Question: > 
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> [$\lim\_{n \to +\infty} n^{\frac{1}{n}} $](https://math.stackexchange.com/questions/115822/lim-n-to-infty-n-frac1n) 
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I know that 

$$\lim\_{n\rightarrow\infty}\sqrt[n]n=1$$

and I can imagine that $n$ grows linearly while $n$th root compresses it exponentially and therefore the result is $1$, but how do I calculate it?

Answer: 
Perhaps one of the most elementary ways to prove it: since $\,n\geq 1\,\,\forall n\in\mathbb{N}\,$ , we can put$$\sqrt[n]{n}=1+c\_n\,,\,c\_n\geq0\Longrightarrow n=(1+c\_n)^n\geq \frac{n(n-1)}{2}c\_n^2$$using the binomial expansion ,so that$$0<c\_n\leq\sqrt\frac{2}{n-1}$$and now just apply the squeeze theorem and get $\,c\_n\to 0\,$, which is precisely what we need.
Let's see a very elementary proof. Without loss of generality we proceed replacing $n$ by $2^n$ and get that:
$$ 1\leq\lim\_{n\rightarrow\infty} n^{\frac{1}{n}}=\lim\_{n\rightarrow\infty} {2^n}^{\frac{1}{{2}^{n}}}=\lim\_{n\rightarrow\infty} {2}^{\frac{n}{{2}^{n}}}\leq\lim\_{n\rightarrow\infty} {2}^{\frac{n}{\dbinom{n}{2}}}=2^0=1$$

By Squeeze Theorem the proof is complete.