Calculating $\lim_{n\to\infty}\sqrt[n]{ \sqrt[n]{n} - 1 }$ I know that   
$$\lim_{n\to\infty} \sqrt[n]{ \sqrt[n]{n} - 1 } = 1,$$
but I'm unable to prove it. I could easily estimate that it's at most $1$, but my best estimation from below is that the limit is greater than $0$.
Doing this from the definition doesn't lead me anywhere either.

We can manipulate limits as long as we are sure that the part we are jumping over is bounded. For example:
$$\lim\limits_{x \to \infty} f(x)^{g(x)} = 
\lim\limits_{x \to \infty} f(x)^{\lim\limits_{x \to \infty}g(x)}$$
as long as $f(x)$ is bounded.
First we attack $\lim\limits_{n \to \infty} n^{\frac{1}{n}}$. We make a substitution $x=\frac{1}{n}$
$$\lim\limits_{n \to \infty} n^{\frac{1}{n}}=\lim\limits_{x \to 0} (\frac{1}{x})^x=\lim\limits_{x \to 0} \frac{1}{x^x}$$
Now this one is elementary $\lim\limits_{x \to 0} x^x = 1$ so we have $\lim\limits_{n \to \infty} n^{\frac{1}{n}}=1$
We continue
$$\lim\limits_{n \to \infty} \sqrt[n]{\sqrt[n]{n}-1}=\lim\limits_{n \to \infty} \sqrt[n]{\sqrt[n]{n}(1-\frac{1}{\sqrt[n]{n}})}=\lim\limits_{n \to \infty}n^{\frac{1}{n^2}} (1-\frac{1}{n^{\frac{1}{n}}})^{\frac{1}{n}}$$
Since $n^{\frac{1}{n^2}}=(n^{\frac{1}{n}})^{\frac{1}{n}}$ we have as well $\lim\limits_{n \to \infty} n^{\frac{1}{n^2}}=1$
That leaves
$$\lim\limits_{n \to \infty} \sqrt[n]{\sqrt[n]{n}-1}=\lim\limits_{n \to \infty} (1-(\frac{1}{n})^{\frac{1}{n}})^{\frac{1}{n}}$$
The same substitution and we have 
$$\lim\limits_{x \to 0} (1-x^x)^{x}$$
$x^x$ has a minimum at $\frac{1}{e}$ and $1 > 1-x^x > 0$ for $0 < x < 1$ meaning it is bounded and $1-x^x$ is not reaching 0 for any $x$ in this region so we can examine
$$\lim\limits_{x \to 0} (1-x^x)^{x}=\lim\limits_{x \to 0} x^{x^2}(\frac{1}{x^x}-1)^{x}=\lim\limits_{x \to 0} (\frac{1}{x^x}-1)^{x}$$
This is because we can write $x^{x^2}=(x^x)^x$
Obviously both limits are equal $\lim\limits_{x \to 0} (\frac{1}{x^x}-1)^{x}=\lim\limits_{x \to 0} (1-x^x)^{x}$.
Notice that
$$\lim\limits_{x \to 0} (\frac{1}{x^x}-1)^{x}=\lim\limits_{x \to 0} (x^{-x}-1)^{x}=\lim\limits_{x \to 0^{-}} (x^{x}-1)^{-x}=\lim\limits_{x \to 0^{-}} \frac{1}{(x^{x}-1)^{x}}$$
which is to say that the right limit is equal to reciprocal left (if it exists). This means that the unique limit, if it exists, is 1 since left and right limits are then the same.
But, $(1-x^x)^x$ is continuous at 0, $x^x$ is continuous and $a^x$ is continuous, meaning the limit is indeed 1.
