Law of Exponent and Telescopic Sum

By the rule of indices, we know that $(x^{m})^{n}=x^{mn}$ . Hence the Given entity can be written in the following form: $A = 32^{\frac{1}{2\times 3}}32^{\frac{1}{3\times 4}}32^{\frac{1}{4\times 5}}...32^{\frac{1}{9\times 10}}$ Again using the fact that $x^{m}x^{n}=x^{m+n}$ , this can be reduced to following form: $A = 32^{x}$ where $x = \frac{1}{2\times 3}+\frac{1}{3\times 4}+\frac{1}{4\times 5}+...+\frac{1}{9\times 10}$

This could be turned into a famous Telescoping sum like this: $x = \left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{5}\right)+...+\left(\frac{1}{9}-\frac{1}{10}\right)$ Cancelling intermediate terms, we get: $x=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}$ Hence we have: $A = 2^{5\times \frac{2}{5}}=\boxed{4}$