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**student2011** Thaaaaaank you very much for your help. I will show you my solution:

So we use induction on the number of prime numbers appears in $\displaystyle a$;

If we have only one prime, then $\displaystyle a_1=p_1^{\alpha_1}$ and the divisors of $\displaystyle a_1$ are $\displaystyle 1,p_1,p_1^{2},....,p_1^{\alpha_1}$. Hence the number of divisors of $\displaystyle a_1$ is $\displaystyle 1+\alpha_1$.

Inductive Step: Suppose that $\displaystyle a_{n-1}$ contains $\displaystyle n-1$ prime numbers, and the statement is true for $\displaystyle a_{n-1}$. i.e $\displaystyle a_{n-1}=p_1^{\alpha_1}p_2^{\alpha_2}....p_{n-1}^{\alpha_{n-1}}$and the number of divisors of $\displaystyle a_{n-1}$ is $\displaystyle \prod_{i=1}^{n-1}{1+\alpha_i}$

Our goal is to show that the statement is true for $\displaystyle a_n$ which contain $\displaystyle n$ prime numbers.

We have $\displaystyle a_n=p_1^{\alpha_1}p_2^{\alpha_2}....p_{n-1}^{\alpha_{n-1}}.p_n^{\alpha_n}$. Hence $\displaystyle a_n=a_{n-1}.p_n^{\alpha_n}$ By inductive hypothesis, the number of divisors of $\displaystyle a_{n-1}$ is $\displaystyle \prod_{i=1}^{n-1}{1+\alpha_i}$ and we know that the number of divisors of $\displaystyle p_n^{\alpha_n}$ is $\displaystyle 1+\alpha_n$. So, the number of divisors of $\displaystyle a_n$ is $\displaystyle \prod_{i=1}^{n-1}{1+\alpha_i}.(1+\alpha_n)$ which is $\displaystyle \prod_{i=1}^{n}{1+\alpha_i$

This is my solution. If there is any mistakes or comments please guide me. Thank you again