A very powerful product

Let $\large S = \displaystyle\prod_{n=0}^{\infty}64^{(\frac{1}{3})^{n}}.$ Then

$\large \log_{64}S = \displaystyle\sum_{n=0}^{\infty} \log_{64}64^{(\frac{1}{3})^{n}} =$ $\displaystyle\sum_{n=0}^{\infty} \left(\dfrac{1}{3}\right)^{n} = \dfrac{1}{1 - \dfrac{1}{3}} = \dfrac{3}{2}.$

Thus $\large S = 64^{\frac{3}{2}} = (2^{6})^{\frac{3}{2}} = 2^{9} = \boxed{512}.$

Start by adding the exponents. $\large \prod _{ n=0 }^{ \infty }{ { 64 }^{ { \left( { 1 }/3 \right) }^{ n } } } ={ 64 }^{ { \left( { 1 }/3 \right) }^{ 0 } }\times { 64 }^{ { \left( { 1 }/3 \right) }^{ 1 } }\times { 64 }^{ { \left( { 1 }/3 \right) }^{ 2 } }\times \dots ={ 64 }^{ 1+\frac { 1 }{ 3 } +\frac { 1 }{ 9 } +\dots }$

The exponent is now a geometric progression with $|r|<1$ , so we can use $\displaystyle {S}_{ \infty } = \frac{a}{1-r}$ . $\large 1+\frac { 1 }{ 3 } +\frac { 1 }{ 9 } +\dots =\sum _{ n=0 }^{ \infty }{ \frac { 1 }{ { 3 }^{ n } } } =\frac { 1 }{ 1-\frac { 1 }{ 3 } } =\frac { 1 }{ \frac { 2 }{ 3 } } =\frac { 3 }{ 2 }$

Finally, $\large \prod _{ n=0 }^{ \infty }{ { 64 }^{ { \left( { 1 }/3 \right) }^{ n } } } ={ 64 }^{ { { 3 }/{ 2 } } } = \color{#20A900}{\boxed{512}}$

Convert to a summation problem by recognizing that exponents add when multiplied.

$S = 1 + \frac{1}{3} + \frac{1}{9} + ...$ $S = 1 + \frac{1}{3}(1 + \frac{1}{3} + \frac{1}{9} + ...)$ $S = 1 + \frac{1}{3} S$ $\frac{2}{3} S = 1$ $S = \frac{3}{2}$

Now evaluate the answer using $\frac{3}{2}$ as your exponent.

$64^{\frac{3}{2}} = \boxed{512}$