Taking an infinite number of square-roots. CORRECTION: Thank you to those who pointed out the mix-up, I meant to write the limit as $i$ approaches infinity.
I came across an interesting application of geometric series. Considering $...\sqrt(\sqrt(\sqrt(\sqrt(\sqrt(...(\sqrt(2)))))) = 2^{\lim_{i \rightarrow \infty}\frac{1}{2^i}} = 2^1 = 2$, it appears to follow that for any arbitrary non-zero, positive, real-number $a$, $a^{\lim_{i \rightarrow \infty}\frac{1}{2^i}} = a$. While I haven't taken complex analysis, is this also true for any complex number? As in, is it true that for non-zero, $a,b \in \mathbb{R}$ that $(a + bi)^{\lim_{i \rightarrow \infty}\frac{1}{2^i}} = a + bi$? 
As an aside, I find the result: $2^{\lim_{i \rightarrow \infty}\frac{1}{2^i}} = 2$ very counter-intuitive. Using a calculator, the more and more square-roots I take of $2$, it appears to get further and further away from 2 $-$ not closer, as I would expect given the result. Any explanation would be appreciated.
 A: You're wrong! Taking "an infinite" number of square roots leads to $$...\sqrt(\sqrt(\sqrt(\sqrt(\sqrt(...(\sqrt(2))))))={\left({\left(2^{\frac{1}{2}}\right)}^{\frac{1}{2}}\right)}^{\frac{1}{2}...}=2^{\frac{1}{2}\frac{1}{2}\frac{1}{2}...}$$
The application would be multiplying an infinite number of square roots of other square roots, i.e.
$$\left(\sqrt{2}\right)\left(\sqrt{\sqrt{2}}\right)\left(\sqrt{\sqrt{\sqrt{2}}}\right)...=2^{\frac{1}{2}}2^{\frac{1}{2^2}}2^{\frac{1}{2^3}}...=2^{\sum \frac{1}{2^i}}=2^1=2$$
And this is true since the function $2^x$ is continuous (so if $a_n$ converges to $a$, $2^{a_n}$ converges to $2^a$). Note this is valid if I change 2 by any complex number.
A: The result you're looking for is this:
$$
\sqrt{a}\cdot\sqrt{\sqrt{a}}\cdot\sqrt{\sqrt{\sqrt{a}}}\cdots = a^{1/2}a^{1/4}a^{1/8}\cdots=a^{1/2+1/4+1/8+\ldots}=a.
$$
When you multiply the different levels of nested square roots together, you get a geometric series summing to $1$ in the exponent.  This will work for complex $a$ as well, as long as you choose a consistent branch for the square roots; for instance, you can lay the branch cut along the negative real axis.
A: Taking an infinite number of square roots like that will not sum any exponents, it will multiply them. You will end up with simply the limit as $n$ approaches infinity of $2^{\frac{1}{2^{n}}}$.
This is $1$, and it is $1$ regardless of whether we have $2^{\frac{1}{2^{n}}}$, $42^{\frac{1}{2^{n}}}$, $i^{\frac{1}{2^{n}}}$, etc.
A: Anyway indeed your question is wrong but $(a+ib)^{1/2^n}\to 1$ as $n \to \infty$
