For $\alpha>1$ find (w/o lhopital)$$ \lim_{n\to\infty} \sqrt[n]{\sqrt[2^n]{\alpha}-1} $$
Progress:
Used this:$$
x^n - y^n = (x-y)(x^{n-1} + x^{n-2} y + ... + x y^{n-2} + y^{n-1})
$$ to get to this:
$$
\alpha-1=(\sqrt[2^n]{\alpha})^{2^n}-1^{2^n}=(\sqrt[2^n]{\alpha}-1)((\sqrt[2^n]{\alpha})^{2^n-1}+(\sqrt[2^n]{\alpha})^{2^n-2}+\dots+\sqrt[2^n]{\alpha}+1)
$$
and so$$
\sqrt[2^n]{\alpha}-1=\frac{\alpha-1}{((\sqrt[2^n]{\alpha})^{2^n-1}+(\sqrt[2^n]{\alpha})^{2^n-2}+\dots+\sqrt[2^n]{\alpha}+1)}
$$
Picking up where the OP left off, there are $2^n$ terms in the sum $(\sqrt[2^n]{\alpha})^{2^n-1}+(\sqrt[2^n]{\alpha})^{2^n-2}+\dots+\sqrt[2^n]{\alpha}+1$, and each is between $1$ and $\alpha$ (since $1\lt\alpha$), so
$$2^n \le(\sqrt[2^n]{\alpha})^{2^n-1}+(\sqrt[2^n]{\alpha})^{2^n-2}+\dots+\sqrt[2^n]{\alpha}+1\le2^n\alpha$$
This implies
$$\lim_{n\to\infty}\sqrt[n]{\sqrt[2^n]\alpha-1}={1\over2}$$
since $\lim_{n\to\infty}\sqrt[n]{\alpha-1}=\lim_{n\to\infty}\sqrt[n]\alpha=1$ when $\alpha\gt1$.
$\sqrt[2^n]{\alpha}=(1+(\alpha -1))^{2^{-n}}=1+(\alpha -1)2^{-n}+o(2^{-n})$
$\sqrt[n]{\sqrt[2^n]{\alpha}-1}=((\alpha -1)2^{-n}+o(2^{-n}))^{\frac{1}{n}}=(\alpha -1)^{\frac{1}{n}}(2^{-n})^{\frac{1}{n}}+o(1)=1\cdot 2^{-1}=\frac{1}{2}$
If you know it is converging, call $u_{n} = \sqrt[2^n]{\alpha}-1$ and note that: $$\lim_{n\to\infty} \sqrt[n]{u_{n}} = \lim_{n\to\infty} \frac{u_{n+1}}{u_{n}} = \lim_{n\to\infty} \frac{1}{\alpha^{\frac{1}{2^{n+1}}}+1} = \frac{1}{2}$$ (The radius of a power series is the same no matter how you compute it).
Let $\epsilon = \alpha - 1>0$ and $a_n={\alpha}^{1/2^n}$. Notice that $a_n>1$ and let $d_n=a_n-1>0$. The expression can then be written as
\begin{align} {d_n}^{1/n} &=\exp\left(\frac1n\,\ln d_n\right).\tag{$*$} \end{align}
We have $\alpha=1+\epsilon=a_n^{2^n}=(1+d_n)^{2^n}=\sum_{k=0}^{2^n}\binom{2^n}k\,d_n^k\,,$ which implies
$$\epsilon=\sum_{k=1}^{2^n}\binom{2^n}k\,d_n^k.$$
In particular, $\epsilon\geq 2^nd_n\iff d_n\leq \epsilon\, 2^{-n}$. It follows that $\ln d_n\leq \ln\epsilon-n\ln2$. Plugging this into $(*)$ yields $d_n^{1/n}\leq \epsilon^{1/n}\,\frac12$.
On the other hand, $1+x\leq \exp x$ for all $x\in\mathbb{R}$ (why?), which implies that $\ln(1+x)\leq x$ whenever $x\geq -1$. It follows that, $\ln\alpha={2^n}\ln(1+d_n)\leq {2^n}d_n$, and hence
$$d_n\geq 2^{-n}\ln\alpha \iff\ln d_n\geq -n\ln 2 +\ln\ln\alpha.$$ Plugging this into $(*)$ yields $d_n^{1/n}\geq {\left(\ln\alpha\right)}^{1/n}\,\frac12$.
We are therefore left with
$${\left(\ln\alpha\right)}^{1/n}\,\frac12\leq d_n^{1/n}\leq \epsilon^{1/n}\,\frac12.$$
Lemma: For each $x>0$ we have $\lim_{n\to\infty}\sqrt[n]{x}=1$.
The proof of the lemma will be left as an exercise. Letting $n\to\infty$, we conclude by the squeeze theoerem together with the lemma that $d_n^{1/n}\to\frac12$.
Barry's answer is clearly more elegant, but I'd already started writing this and I figure the bounds can be at least of some interest for estimating speed of convergence.
