Let $\displaystyle d_n = (2^{n!} - 1, m)$

Explain why $\displaystyle d_n \mid d_{n+1}$ for n = 1,2,...

Show that $\displaystyle d_n > 1$ if m has a prime factor p such that $\displaystyle (p-1) \mid n!$.

My idea is to show that $\displaystyle (2^{n!}-1,m)$ can only equal factors of m, and since $\displaystyle d_n and d_{n+1}$ have the same m in the gcd and that $\displaystyle d_n < d_{n+1}$, $\displaystyle d_n \mid d_{n+1}$.

I don't think this reasoning is rigorous.

Also, I have trouble with showing that $\displaystyle d_n > 1$ if m has a prime factor p such that $\displaystyle (p-1) \mid n!$

Thanks!