Let $m_i\geq 2 (1\leq i\leq n)$ be $n$ pairwisely coprime positive integers and let $q_i\geq 2 (1\leq i\leq n)$ be $n$ arbitrary prime powers, let$A=\prod_{i=1}^n(({q_i}^{m_i}1)/(q_i1))$. Let $\sigma(A)$ be the number of different prime factors of A, is it true that $\sigma(A)\geq n$? If this is not true, is there a counterexample? Is there a good way to estimate $\sigma(A)$?

No it's not true you have the following counterexample: $$ \frac{2^51}{21} \times \frac{5^31}{51} = 31^2 $$ 


Let take a look to the special case when your $q_i$ are actually $n$ distinct \emph{odd} prime numbers. I use the standard notations : $\omega(H)$ is the number of distinct prime divisors of $H$ and $\sigma(G)$ is the sum of all positive divisors of $G.$ Put $B$ the product of all the $q_i^{m_i}$ then we have $$ \sigma(B) = mB $$ if $B$ is an odd $m$multiperfect number. ((sure, we do not known concrete examples of this, but...) So, in this case $$ \omega(B) = n $$ and you have your lower bound attained. luis 

