7.009. $\left(N^{\frac{1}{\log _{2} N}} \cdot N^{\frac{1}{\log _{4} N}} \cdot N^{\frac{1}{\log _{8} N}} \cdots N^{\frac{1}{\log _{64} N}}\right)^{\frac{1}{15}}$ (the bases of the logarithms are consecutive natural powers of the number 2).
## Solution.

$$
\begin{aligned}
& \left(N^{\frac{1}{\log _{2} N}} \cdot N^{\frac{1}{\log _{4} N}} \cdot N^{\frac{1}{\log _{8} N}} \cdots N^{\frac{1}{\log _{512} N}}\right)^{\frac{1}{15}}= \\
& =\left(N^{\log _{N} 2} \cdot N^{\log _{N} 4} \cdot N^{\log _{N} 8} \cdots N^{\log _{N} 512}\right)^{\frac{1}{15}}= \\
& =(2 \cdot 4 \cdot 8 \cdots 512)^{\frac{1}{15}}=\left(2^{1} \cdot 2^{2} \cdot 2^{3} \cdots 2^{9}\right)^{\frac{1}{15}}=\left(2^{1+2+3+\ldots+9}\right)^{\frac{1}{15}}
\end{aligned}
$$

The expression $S_{n}=1+2+3+\ldots+9$ is the sum of the terms of an arithmetic progression, where $a_{1}=1, d=1, a_{n}=9, n=9$. Then $S_{n}=\frac{a_{1}+a_{n}}{2} n=$ $=\frac{1+9}{2} \cdot 9=45$. Therefore, $\left(2^{45}\right)^{\frac{1}{15}}=2^{3}=8$.

Answer: 8.