Why is $x^n\approx \left(n(x^{1/4096}-1)+1\right)^{4096}$? There's an old-school pocket calculator trick to calculate $x^n$ on a pocket calculator, where both, $x$ and $n$ are real numbers. So, things like $\,0.751^{3.2131}$ can be calculated, which is awesome.
This provides endless possibilities, including calculating nth roots on a simple pocket calculator.
The trick goes like this:

*

*Type $x$ in the calculator

*Take the square root twelve times

*Subtract one

*Multiply by $n$

*Add one

*Raise the number to the 2nd power twelve times (press * and = key eleven times)

Example:
I want to calculate $\sqrt[3]{20}$ which is the same as $20^{1/3}$. So $x=20$ and $n=0.3333333$. After each of the six steps, the display on the calculator will look like this:

*

*$\;\;\;20$

*$\;\;\;1.0007315$

*$\;\;\;0.0007315$

*$\;\;\;0.0002438$

*$\;\;\;1.0002438$

*$\;\;\;2.7136203$
The actual answer is $20^{1/3}\approx2.7144176$. So, our trick worked to three significant figures. It's not perfect, because of the errors accumulated from the calculator's 8 digit limit, but it's good enough for most situations.

Question:
So the question is now, why does this trick work? More specifically, how do we prove that:
$$x^n\approx \Big(n(x^{1/4096}-1)+1\Big)^{4096}$$

Note: $4096=2^{12}$.
I sat in front of a piece of paper trying to manipulate the expression in different ways, but got nowhere.
I also noticed that if we take the square root in step 1 more than twelve times, but on a better-precision calculator, and respectively square the number more than twelve times in the sixth step, then the result tends to the actual value we are trying to get, i.e.:

$$\lim_{a\to\infty}\Big(n(x^{1/2^a}-1)+1\Big)^{(2^a)}=x^n$$

This, of course doesn't mean that doing this more times is encouraged on a pocket calculator, because the error from the limited precision propagates with every operation. $a=12$ is found to be the optimal value for most calculations of this type i.e. the best possible answer taking all errors into consideration. Even though 12 is the optimal value on a pocket calculator, taking the limit with $a\to\infty$ can be useful in proving why this trick works, however I still can't of think a formal proof for this.
Thank you for your time :)
 A: For fixed $x > 0$ and $n$, let $t = 1/2^a \to 0$. Then we need to prove that
$$
 \lim_{t \to 0} \left( n (x^t - 1) + 1 \right)^{1/t} = x^n.
$$
In fact, we have
$$
 \ln \left[\lim_{t \to 0} \left( n (x^t - 1) + 1 \right)^{1/t}\right] = 
\lim_{t \to 0} \frac{\ln (1 + n(x^t - 1))}{t} = \lim_{t \to 0} \frac{n(x^t - 1)}{t} = n\ln x,
$$
where the first equality follows from the continuity of $\ln(x)$, and the second equality has used the fact that $\ln(1 + x) \sim x$ when $x \to 0$.
A: A standard trick is to calculate the natural logarithm first to get the exponent under control:
$$\log(\lim_{a\to\infty}(n(x^{1/a}-1)+1)^a)=\lim_{a\to\infty}a\log(nx^{1/a}-n+1)$$
Set $u=1/a$.
We get
$$\lim_{u\to 0}\frac{\log (nx^u-n+1)}{u}$$
Use L'Hopital:
$$\lim_{u\to 0}\frac{nx^u\log x}{nx^u-n+1}=n\log x=\log x^n$$
Here we just plugged in $u=0$ to calculate the limit!
So the original limit goes to $x^n$ as desired.
A: If $x$ (actually $\ln x$) is relatively small, then $x^{1/4096}=e^{(\ln x)/4096}\approx1+(\ln x)/4096$, in which case
$$n(x^{1/4096}-1)+1)\approx1+{n\ln x\over4096}$$
If $n$ is also relatively small, then
$$(n(x^{1/4096}-1)+1)^{4096}\approx\left(1+{n\ln x\over4096}\right)^{4096}\approx e^{n\ln x}=x^n$$
Remark: When I carried out the OP's procedure on a pocket calculator, I got the same approximation as the the OP, $2.7136203$, which is less than the exact value, $20^{1/3}=2.7144176\ldots$. Curiously, the exact value (according to Wolfram Alpha) for the approximating formula,
$$\left({1\over3}(20^{1/4096}-1)+1\right)^{4096}=2.7150785662\ldots$$
is more than the exact value. On the other hand, if you take the square root of $x=20$ eleven times instead of twelve -- i.e., if you use $2048$ instead of $4096$ -- the calculator gives $2.7152613$ while WA gives $2.715739784\ldots$, both of which are too large.  Very curiously, if you average the two calculator results, you get
$${2.7136203+2.7152613\over2}=2.7144408$$
which is quite close to the true value!
