Twisted Nested Radicals!

$P=2\sqrt{2\sqrt[3] {2\sqrt[4]{2\sqrt[5] {2\cdots}}}}$

The above expression can be rewritten as ,

$=2(2(2(2(2...)^{\frac{1}{5}})^{\frac{1}{4}})^{\frac{1}{3}})^{\frac{1}{2}}$

$=2\cdot 2^{\frac{1}{2}}\cdot 2^{\frac{1}{6}}\cdot 2^{\frac{1}{24}}\cdot 2^{\frac{1}{120}}...$

$=2^{1+\frac{1} {2!}+\frac{1}{3!}+\frac{1} {4!}+\frac{1}{5!}+\cdots}$

$=2^{e-1}=\frac{2^{e}}{2}$

$\Rightarrow 2+2=\boxed{4}$

$P=2\sqrt{2\sqrt[3]{2\sqrt[4]{2\sqrt[5]{2\ldots}}}}=2(2^{\frac 1 {2!}}(2^{\frac 1 {3!}}(2^{\frac 1 {4!}}(2^{\frac 1 {5!}}\ldots))))$

$P=2^{{{\sum}_{k=0}^\infty}(\frac 1 {k!})-1}=2^{e-1}=\frac {2^e} 2 =\frac {A^e} B$

$A+B=2+2=\boxed 4$

$P = 2^{\frac{1}{1!}} 2^{\frac{1}{2!}} ... = 2^{ \frac{1}{1!} + \frac{1}{2!} + ... } = 2^{e-1}= \frac{2^e}{2}$

$A = 2, B = 2 \Rightarrow A + B =\boxed{4}$

Not a rigorous proof, but from the given value of P ( some power 2 ), you can guess from the equality that A^e is also some power of 2, and B is just 2. If there were any other numbers added to the equality, guessing would not work out so well