Alternate method to solve $\sqrt{11\sqrt{11\sqrt{11...4\, \text{times}}}}$

Question

Question :

What is the value of $$\sqrt{11\sqrt{11\sqrt{11...4\,\text{times}}}}$$

I did it by solving square root one by one. $$\sqrt{11\sqrt{11\sqrt{11\times11^\frac{1}{2}}}}$$ $$\sqrt{11\sqrt{11\sqrt{11^\frac{3}{2}}}}$$ $$\sqrt{11\sqrt{11\times{11^\frac{3}{4}}}}$$ $$\sqrt{11\sqrt{11^\frac{7}{4}}}$$ $$\sqrt{11\times{11^\frac{7}{8}}}$$ $$\sqrt{11^\frac{15}{8}}$$ $$11^\frac{15}{16}$$

Is there any other way to solve this?

I don't want the complete solution, just tell me the approach.

Answer 1

If you solve the square roots one by one, but from outer one to inner one, you see it is the sum of geometric sequence.

Answer 2

There is a faster way to solve it, actually:

$\sqrt{11·\sqrt{11·\sqrt{11·\sqrt{11·\color{blue}{11}}}}} = \sqrt{11·\sqrt{11·\sqrt{11·11}}} = \sqrt{11·\sqrt{11·11}} = \sqrt{11·11} = 11$

So clearly the extra factor is just $\sqrt{\sqrt{\sqrt{\sqrt{\color{blue}{11}}}}}$, since square-root commutes with multiplication on non-negative reals (which implies that in any expression comprising just multiplication and square-roots the number of square-roots in effect is just the number of 'radical lines' above it).