A long time ago I solved the following theorem
Let $p_1,p_2,\ldots,p_k$ be distinct primes. Let $\{a_i\}^\infty_{i=1}$ be the increasing sequence of positive integers whose prime factorization contains only these primes (not necesarily all). Show that $\forall c>0\exists n\in \mathbb{N}:a_{n+1}-a_n>c$
Solution. Let $m$ be an integer such that $p^m_i>c$ for all $i=1,2,\ldots,k$. Let $a_n=(a_1a_2\ldots a_k)^m$. As $a_{n+1}>a_n$, there exist a prime $p_j$ such that $p_j^m|a_{n+1}$. Then $p^m|a_{n+1}-a_n$ from which $a_{n+1}-a_n>c$.
Then a friend told me that this stronger version is also true.
Let $p_1,p_2,\ldots,p_k$ be distinct primes. Let $\{a_i\}^\infty_{i=1}$ be the increasing sequence of positive integers whose prime factorization contains only these primes (not necesarily all). Show that $\forall c>0\exists n_c\in \mathbb{N}:n>n_c\implies a_{n+1}- a_n>c$
I've tried to solve it but I couldn't. I'm 90% sure that the stronger version is also true. I'm interested in elementary solutions but a complex one is also welcome.