5 If for a positive integer $n$, there exists $k$, such that
$$
n=n_{1} n_{2} \cdots n_{k}=2^{\frac{1}{k^{k}\left(n_{1}-1\right) \cdots\left(n_{k}-1\right)}-1}
$$

where $n_{1}, \cdots, n_{k}$ are all integers greater than 3, then $n$ is said to have property $P$. Find all numbers $n$ that have property $P$.
From the condition, if $n$ has property $P$, then $n=2^{m}-1$, where $m$ is a positive integer.

Verification shows that for $m=1,2, \cdots, 9$, only when $m=3$, $2^{m}-1$ has property $P$, at this time $n=7$.
Below, we prove that for $m \geqslant 10, n=2^{m}-1$ does not have property $P$. We use proof by contradiction. Assume there exists some $m \geqslant 10$ and a positive integer $k$, such that $n=2^{m}-1$ can be expressed as
$$
n=n_{1} n_{2} \cdots n_{k}=2^{\frac{1}{k}\left(n_{1}-1\right) \cdots\left(n_{t}-1\right)}-1
$$

Then we have
$$
\frac{1}{2^{k}}\left(n_{1}-1\right) \cdots\left(n_{k}-1\right) \geqslant 10
$$

By induction, it can be proven: when $l \geqslant 10$, $2^{l}-1>l^{3}$. Therefore,
$$
22^{\frac{1}{k}\left(n_{1}-1\right) \cdots\left(n_{k}-1\right)}-1>\left(\frac{1}{2^{k}}\left(n_{1}-1\right) \cdots\left(n_{k}-1\right)\right)^{3}
$$

Since $n=2^{m}-1$, $n_{i}$ are all odd numbers. Also, $n_{i}>3$, so $n_{i} \geqslant 5(1 \leqslant$ $i \leqslant k)$. Thus,
$$
\left(\frac{n_{i}-1}{2}\right)^{3} \geqslant 4 \cdot \frac{n_{i}-1}{2}>n_{i}, 1 \leqslant i \leqslant k
$$

Therefore, 
$$
2^{\frac{1}{2^{k}\left(n_{1}-1\right) \cdots\left(n_{k}-1\right)}}-1>\left(\frac{1}{2^{k}}\left(n_{1}-1\right) \cdots\left(n_{k}-1\right)\right)^{3}>
$$
$$
n_{1} n_{2} \cdots n_{k}
$$

This contradicts equation (1).
Therefore, the number with property $P$ is only $n=7$.