Lots of 32

$\begin{aligned} \prod_{n=1}^8 {\left[ \left( 32 \right)^{\frac{1}{n+1}} \right]^{\frac{1}{n+2}}} & = \prod_{n=1}^8 {\left[ \left( 2^5 \right) ^{\frac {1} {n+1}} \right]^{\frac{1}{n+2}}} = \prod_{n=1}^8 {\left[2^{\frac {5} {n+1}} \right]^{\frac{1}{n+2}}} = \prod_{n=1}^8 {2^{\frac {5} {(n+1)(n+2)}}} \\ & = 2^{\sum_{n=1}^8 {\frac {5} {(n+1)(n+2)}}} = 2^{5 \sum_{n=1}^8 {\left( \frac {1} {n+1} - \frac {1} {n+2}\right)}} \\ & = 2^{5 \left( \frac{1}{2} - \frac{1}{10} \right)} = 2^2 = \boxed{4} \end{aligned}$

$\LARGE \sqrt[3]{ \sqrt{32}} \sqrt[4]{ \sqrt[3]{32}} \sqrt[5]{ \sqrt[4]{32}} \cdots \sqrt[10]{ \sqrt[9]{32}}$ $\LARGE 32^{\frac{1}{2} \times \frac{1}{3}} 32^{\frac{1}{3} \times \frac{1}{4}} 32^{\frac{1}{4} \times \frac{1}{5}} ... 32^{\frac{1}{9} \times \frac{1}{10}}$ $\LARGE 32^{\frac {1}{2.3} + \frac {1}{3.4} + \frac {1}{4.5} +... +\frac{1}{9.10}}$ $\LARGE 32^{\frac{2}{5}}$ $\LARGE \boxed{4}$

The fractional exponents are: $\frac{1}{3}*\frac{1}{2},\frac{1}{4}*\frac{1}{3},...\frac{1}{n}*\frac{1}{n-1},...\frac{1}{10}*\frac{1}{9}$ .

This makes the series: $\prod_{n=3}^{10}(32^\frac{1}{n(n-1)})$

Because $a^m*a^n=a^{mn}$ , this in turn is: $\prod_{n=3}^{10}(32^\frac{1}{n(n-1)}) = 32^{\sum_{n=3}^{10}(\frac{1}{n(n-1)})}$

The series $\sum_{n=2}^{m}(\frac{1}{n(n-1)}) = \frac{m-1}{m}$ , but since we're starting at $n=3$ , we'll need to subtract the value where $n=2$ . (A quick substitution shows this to be $\frac{1}{2}$ .)

So: $\sum_{n=3}^{m}(\frac{1}{n(n-1)}) = \sum_{n=2}^{m}(\frac{1}{n(n-1)}) - \frac{1}{2} = \frac{10-1}{10} - \frac{1}{2} \\ = \frac{9}{10} - \frac{1}{2} = \frac{4}{10} = \frac{2}{5}$

Thus: $\prod_{n=3}^{10}(32^\frac{1}{n(n-1)}) = 32^{\sum_{n=3}^{10}(\frac{1}{n(n-1)})} = 32^{\frac{2}{5}} = {2^5}^{\frac{2}{5}} = 2^2 = 4$

Sum of the exponents of 32 are $\frac {1}{2.3} + \frac {1}{3.4} + \frac {1}{4.5} +...... +\frac{1}{9.10}$ which adds up to $\frac{1}{2} - \frac{1}{10}$ = $\frac {2}{5}$ as exponent of 32. Answer = 4

It goes like that: A=sqrt of 3(sqrt(32)) sqrt of 4(sqrt of 3(32)) ...* sqrt of 10(sqrt of 9(32)) =(32^(1/2 * 1/3)) (32^(1/3 * 1/4)) ...*(32^(1/9 * 1/10)) =32^( 1/2 * 1/3 + 1/3 * 1/4 +...+ 1/9 * 1/10 )=32^X

X=1/2 * 1/3 + 1/3 * 1/4 + ... + 1/9 * 1/10 =1/2 - 1/3 + 1/3 - 1/4 + ... + 1/9 - 1/10 =1/2 - 1/10 = 2/5

A=32^(2/5)=4 Nice one.

32^(1/2.3+1/3.4+1/4.5+.................+1/9.10) 1/2.3=1/2-1/3 1/3.4=1/3-1/4 so (1/2-1/3)+(1/3-1/4)+(1/4-1/5).................(1/9-1/10) =1/2-1/10 =====>2/5 32^(2/5) =======>((32)^1/5)^2===========>2^2=4