Why does $\lim_{n \to \infty}(1+\frac{1}{4^n})^{2^n} $ not equal $\sqrt{e}$? I tried to evaluate $\lim_{n \to \infty}(1+\frac{1}{4^n})^{2^n}$ with
$$\lim_{n \to \infty}(1+\frac{1}{4^n})^{2^n} = \lim_{n \to \infty}(1+\frac{1}{4^n})^{(4^\frac{1}{2})^n} = \lim_{n \to \infty}(1+\frac{1}{4^n})^{4^\frac{n}{2}} = \lim_{n \to \infty}((1+\frac{1}{4^n})^{4^n})^\frac{1}{2} = ({e})^\frac{1}{2} = \sqrt{e}.$$
However, I know that the correct answer is $\lim_{n \to \infty}(1+\frac{1}{4^n})^{2^n} = 1$. So at what point did I make a mistake? Which move was not allowed?
 A: Note that in general $\sqrt{A^x} = A^{\frac{x}{2}} \not = A^{\sqrt{x}}$ [as in general $\frac{x}{2} \not = \sqrt{x}$]. Try $A=2$ and $x=9$ to see for yourself.
In particular here: $$\left(1+\frac{1}{4^n}\right)^{2^n} = \left(1+\frac{1}{4^n}\right)^{\sqrt{4^n}} \not = \sqrt{\left(1+\frac{1}{4^n}\right)^{4^n}}.$$
Instead:
$$\left(1+\frac{1}{4^n}\right)^{2^n} = \left(\left(1+\frac{1}{4^n}\right)^{4^n}\right)^{2^{-n}},$$ and as the inequality $4 \ge \left(1+\frac{1}{4^n}\right)^{4^n} \ge 1$ holds for each $n$, it follows that, on the one hand:
$$\lim_{n \rightarrow \infty} \left(1+\frac{1}{4^n}\right)^{2^n} = \lim_{n \rightarrow \infty}\left(\left(1+\frac{1}{4^n}\right)^{4^n}\right)^{2^{-n}}$$
$$\le \lim_{n \rightarrow \infty} 4^{2^{-n}} = 1.$$
And on the other hand:
$$\lim_{n \rightarrow \infty} \left(1+\frac{1}{4^n}\right)^{2^n} = \lim_{n \rightarrow \infty}\left(\left(1+\frac{1}{4^n}\right)^{4^n}\right)^{2^{-n}}$$
$$\ge \lim_{n \rightarrow \infty} 1^{2^{-n}} = 1.$$
Thus indeed,
$$\lim_{n \rightarrow \infty} \left(1+\frac{1}{4^n}\right)^{2^n}  = 1.$$
A: Here is slightly different argument.
A well known result in Calculus states that

If $a_n$ is a numeric sequence (real or complex) and that $a_n\xrightarrow{n\rightarrow\infty}a$, then
$$\lim_{n\rightarrow\infty}\Big(1+\frac{a_n}{n}\Big)^n=e^{a}$$

This result has been discussed before in MSE in other contexts, see for for example here.
Now,
$$A_n:=\Big(1+\frac{1}{4^n}\Big)^{2^n}=\Big(1+\frac{2^{-n}}{2^n}\Big)^{2^n}$$
Then, as from the result mentioned above, the sequence $A_n$ converges to $e^{\lim_n 2^{-n}}=e^0=1$.
