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The product is equal to $4^{1/3} = \sqrt[3]{4}$, so a + b = 7.

Let $P = 4^{1/4} \cdot 16^{1/16} \cdot 64^{1/64} \cdot 256^{1/256} \dotsm$
Then $P$ can be expressed in the form $\sqrt[a]{b}$,
where $a$ and $b$ are positive integers.
Find the smallest possible value of $a+b$

$\begin{array}{|rcll|} \hline P &=& 4^{1/4} \cdot 16^{1/16} \cdot 64^{1/64} \cdot 256^{1/256} \dotsm \\ P &=& 2^{\frac{1}{2^1}} \cdot 2^{\frac{2}{2^3}} \cdot 2^{\frac{3}{2^5}} \cdot 2^{\frac{4}{2^7}} \dotsm 2^{\left(\frac{n}{2^{2n-1}} \right)} \dotsm \\ P &=& 2^{\frac{1}{2^1}+\frac{2}{2^3}+\frac{3}{2^5}+\frac{4}{2^7}+\dotsm+\frac{n}{2^{2n-1}}+\dotsm } \\ P &=& 2^{S_{\infty}} \\ \hline S_n &=& \frac{1}{2^1}+\frac{2}{2^3}+\frac{3}{2^5}+\frac{4}{2^7}+\dotsm+\frac{n}{2^{2n-1}} \\ \frac{1}{2^2}S_n &=& \frac{1}{2^3}+\frac{2}{2^5}+\frac{3}{2^7}+\dotsm+\frac{n-1}{2^{2n-1}} +\frac{n}{2^{2n+1}}\\ \hline S_n-\frac{1}{2^2}S_n &=& \frac{1}{2^1}+\frac{1}{2^3}+\frac{1}{2^5}+\frac{1}{2^7}+\dotsm+\frac{1}{2^{2n-1}}-\frac{n}{2^{2n+1}} \\ \frac{3}{4}S_n &=& s-\frac{n}{2^{2n+1}}\qquad (1) \\ s&=& \frac{1}{2^1}+\frac{1}{2^3}+\frac{1}{2^5}+\frac{1}{2^7}+\dotsm+\frac{1}{2^{2n-1}} \\ \frac{1}{2^2}s &=& \frac{1}{2^3}+\frac{1}{2^5}+\frac{1}{2^7}+\dotsm+\frac{1}{2^{2n-1}}+\frac{1}{2^{2n+1}} \\ \hline s-\frac{1}{2^2}s &=&\frac{1}{2^1} - \frac{1}{2^{2n+1}} \\ \frac{3}{4}s &=&\frac{1}{2^1} - \frac{1}{2^{2n+1}} \\\\ s &=& \frac{4}{3}* \left( \frac{1}{2^1} - \frac{1}{2^{2n+1}} \right) \\ \hline \frac{3}{4}S_n &=& s-\frac{n}{2^{2n+1}}\qquad (1) \\ \frac{3}{4}S_n &=& \frac{4}{3}* \left( \frac{1}{2^1} - \frac{1}{2^{2n+1}} \right)-\frac{n}{2^{2n+1}} \\ \frac{3}{4}S_{\infty} &=& \frac{4}{3}* \left( \frac{1}{2^1} - 0 \right)-0 \\ \frac{3}{4}S_{\infty} &=& \frac{2}{3} \\ S_{\infty} &=& \frac{4}{3}*\frac{2}{3} \\ S_{\infty} &=& \frac{8}{9} \\ \hline P &=& 2^{S_{\infty}} \\ P &=& 2^{\frac{8}{9}} \\ P &=& \sqrt[9]{2^8} \\ P &=& \sqrt[9]{256} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline a+b &=& 9+256 \\ \mathbf{a+b} &=& \mathbf{265} \\ \hline \end{array}$