2. For integers $n \geq 3$, define the sequence $\alpha_{1}, \alpha_{2}, \mathrm{~L}, \alpha_{k}$ as the exponents corresponding to the primes $p_{1}<p_{2}<\mathrm{L}<p_{k}$ in the prime factorization of $n!$, $n!=p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \mathrm{~L} p_{k}^{\alpha_{k}}$. Find all integers $n \geq 3$ such that $\alpha_{1}, \alpha_{2}, \mathrm{~L}, \alpha_{k}$ form a geometric sequence.
II. Analysis: We call the numbers that satisfy the conditions in the problem as good numbers. Then we have the following lemma:

Lemma: Given primes $p_{1}, p_{2}$ satisfying $p_{1}\sum_{t=1}^{\left[\log _{3} n\right]}\left(\frac{n}{3^{t}}-1\right)=\frac{n}{2}-\left[\log _{3} n\right]-\frac{n}{2 \times 3^{\left[\log _{3} n\right]}}>\frac{n}{2}-\log _{3} n-\frac{3}{2}$

Similarly, $\alpha_{4}>\frac{n}{6}-\log _{7} n-\frac{7}{6}, \alpha_{3}=\sum_{t=1}^{+\infty}\left[\frac{n}{5^{t}}\right] \leq \sum_{t=1}^{+\infty} \frac{n}{5^{t}}=\frac{n}{4}$;

From $\alpha_{3}^{2}=\alpha_{4} \alpha_{2}$, we know: $\left(\frac{n}{4}\right)^{2}>\left(\frac{n}{2}-\log _{3} n-\frac{3}{2}\right)\left(\frac{n}{6}-\log _{7} n-\frac{7}{6}\right)$;

That is, $\frac{3}{4}>\left(1-\frac{2 \log _{3} n}{n}-\frac{3}{n}\right)\left(1-\frac{6 \log _{7} n}{n}-\frac{7}{n}\right) \quad(*)$.
When $n \geq 200$, $\frac{\log _{3} n}{n} \leq \frac{\log _{3} 200}{200}\left(1-\frac{1}{20}-\frac{3}{200}\right)\left(1-\frac{9}{100}-\frac{7}{200}\right)>\frac{3}{4}$, which is a contradiction. Therefore, $n \leq 200$.

Taking $p_{1}=101, p_{2}=103$, by the lemma, we have: $n \leq 102$; taking $p_{1}=53, p_{2}=59$, by the lemma, we have: $n \leq 58$;

Taking $p_{1}=31, p_{2}=37$, by the lemma, we have: $n \leq 36$; taking $p_{1}=19, p_{2}=23$, by the lemma, we have: $n \leq 22$;

Taking $p_{1}=13, p_{2}=17$, by the lemma, we have: $n \leq 16$; taking $p_{1}=11, p_{2}=13$, by the lemma, we have: $n \leq 12$;

Therefore, $7 \leq n \leq 12$, and upon inspection, only $n=10$ meets the conditions;

In summary, all $n$ are $3,4,6,10$.