34. Find the product of all divisors of the number $N$, if there are $n$ of them.
34. Let the divisors of the number $N, a_{1}, a_{2}, a_{3}, \ldots, a_{n}$ form an increasing sequence. Then $a_{1}=1, a_{n}=N$. If $n$ is even, then $a_{i} a_{\gamma}=N$ (there will be pairs equidistant from the beginning and end), then $N \cdot N \cdot N \ldots N=\frac{n}{N^{2}}$ (a total of $\frac{n}{2}$ pairs). If $n$ is odd, then there will be $\frac{n-1}{2}$ pairs, the middle divisor $\sqrt{N}$, the product of divisors $N^{\frac{n-1}{2}} N^{\frac{1}{2}}=N^{\frac{n}{2}}$. Answer: the product of divisors is $N^{\frac{n}{2}}$.