Why does $a=\prod_{n=1}^{∞}\left(a^{\frac{1}{2^{n}}}\right)$ So before I start, disclaimer that I'm just a high school student, so apologies if I'm missing something obvious here.
A while back I found the expression
$$\frac{\sqrt{n\sqrt{n\sqrt{n\sqrt{n\sqrt{n\sqrt{n}}}}}}}{n}$$
which led me to start experimenting with
$$\prod_{n=1}^{∞}\left(a^{\frac{1}{2^{n}}}\right)$$
and I found that when graphed out in the form
$$y=\prod_{n=1}^{x}\left(a^{\frac{1}{2^{n}}}\right)$$
it always tended towards $y = a$, and I don't quite understand why, and was hoping someone would explain.
 A: Informally:
$$\prod_{n=1}^\infty a^{\frac{1}{2^n}} = a^{\sum_{n=1}^\infty \frac{1}{2^n}} = a^1=a.$$

More formally, you can observe that
$$\prod_{n=1}^N a^{\frac{1}{2^n}} = a^{\sum_{n=1}^N \frac{1}{2^n}} = a^{1 - \frac{1}{2^{N+1}}} = a\cdot a^{-\frac{1}{2^{N+1}}}$$
and in the limit, you get
$$\prod_{n=1}^\infty a^{\frac{1}{2^n}} = \lim_{N\to\infty}\prod_{n=1}^N a^{\frac{1} {2^n}} = \lim_{N\to\infty} a\cdot a^{-\frac{1}{2^{N+1}}} = a\cdot \lim_{N\to\infty} a^{-\frac{1}{2^{N+1}}} = a\cdot 1 = a.$$
Note that the penultimate equality above uses quite a bit of calculus, and can be boiled down to:

*

*No matter what $a$ is, you have $a^0 = 1$.

*The function $x \mapsto a^x$ is continuous.

*If $f$ is a continuous function and $x_n$ is a convergent sequence, then $\lim_{n\to\infty} f(x_n) = f(\lim_{n\to\infty} x_n)$.

*Therefore, from 2 and 3, you have $\lim_{N\to\infty} a^{-\frac{1}{2^{N+1}}} = a^{\lim_{N\to\infty}-\frac{1}{2^{N+1}}} = a^0 = 1.$

With even more mathematical firepower, you could use the theorem that, if $a_n>0$ for all $n$, tells you that $$\prod_{n=1}^\infty a_n$$ converges if and only if, for any $b>0$, the sum $$\sum_{n=1}^\infty \log_b a_n$$ converges. The theorem also tells you that if the two converge, then $$\prod_{n=1}^\infty a_n=b^{\sum_{n=1}^\infty \log_b a_n},$$
so you can quickly see this is directly applicable to your problem. Taking $a_n = a^{\frac{1}{2^n}}$ and $b=a$, the theorem tells you that
$$\prod_{n=1}^\infty a^{\frac{1}{2^n}} = a^{\sum_{n=1}^\infty} \log_a\left( a^{\frac{1}{2^n}}\right) = a^{\sum_{n=1}^\infty \frac{1}{2^n}} = a^1 =a$$
A: Being inexact:
$$\prod_{n=1}^{∞}\left(a^{\frac{1}{2^{n}}}\right) = a^{\sum_{n=1}^{∞}\left(\frac{1}{2^{n}}\right)}= a^1 = a $$
So if the infinite product converges, it must converge to $a$
