8. For integer $n \geqslant 3$, let $n!=\prod_{i=1}^{k} p_{i}^{\alpha_{i}}$, where $p_{1}, p_{2}, \cdots, p_{k}$ are all prime numbers, and $p_{1}<p_{2}<\cdots<p_{k}$, $\alpha_{1}, \alpha_{2}, \cdots, \alpha_{k}$ are all positive integers. Find all $n(n \geqslant$ $3)$ such that $\alpha_{1}, \alpha_{2}, \cdots, \alpha_{k}$ form a geometric sequence.
8. Let $[x]$ denote the greatest integer not exceeding the real number $x$.

When $n \geqslant 11$, if $n$ satisfies the condition, in the standard factorization of $n!$,
the exponent of 3 is
$$
\begin{array}{l}
\alpha_{2}=\sum_{k=1}^{\infty}\left[\frac{n}{3^{k}}\right] \geqslant\left[\frac{n}{3}\right]+\left[\frac{n}{9}\right] \\
\geqslant \frac{n-2}{3}+\frac{n-8}{9}=\frac{4 n-14}{9} ;
\end{array}
$$
the exponent of 5 is
$$
\alpha_{3}=\sum_{k=1}^{\infty}\left[\frac{n}{5^{k}}\right]<\sum_{k=1}^{\infty} \frac{n}{5^{k}}=\frac{n}{4} ;
$$
the exponent of 7 is
$$
\alpha_{4}=\sum_{k=1}^{\infty}\left[\frac{n}{7^{k}}\right] \geqslant\left[\frac{n}{7}\right] \geqslant \frac{n-6}{7} .
$$

Since $\alpha_{2} \alpha_{4}=\alpha_{3}^{2}$, we have
$$
\frac{4 n-14}{9} \cdot \frac{n-6}{7}<\frac{n^{2}}{16} \Rightarrow n \leqslant 605 \text {. }
$$

For prime numbers $p$ and $q$, if $p<q<2 p$, then when $q \leqslant n \leqslant 2 p-1$, in the standard factorization of $n!$, the exponents of $p$ and $q$ are both 1. However, the exponents of $p$ and $q$ are two different terms in a sequence of positive integers with a common ratio greater than 1, which cannot be equal, leading to a contradiction.

Thus, there is no positive integer $n$ that satisfies the condition in the interval $[q, 2 p-1]$.
Notice, Table 1.
Table 1
\begin{tabular}{|c|c|c|c|c|c|c|c|c|}
\hline$p$ & 7 & 11 & 17 & 29 & 47 & 83 & 157 & 311 \\
\hline$q$ & 11 & 13 & 19 & 31 & 53 & 89 & 163 & 313 \\
\hline $2 p-1$ & 13 & 21 & 33 & 57 & 93 & 165 & 313 & 621 \\
\hline
\end{tabular}

Then, in the eight closed intervals $[11,13]$, $[13,21]$, $[19,33]$, $[31,57]$, $[53,93]$, $[89,165]$, $[163,313]$, and $[313,621]$, there is no positive integer $n$ that satisfies the condition. The union of these eight intervals is $[11,621]$.
Therefore, when $11 \leqslant n \leqslant 605$, $n$ does not satisfy the condition.
For $n (3 \leqslant n \leqslant 10)$, it can be verified that $n=3,4,6,10$ satisfy the condition.
Thus, $n=\{3,4,6,10\}$.