Simplifying $\text{lcm}(a^{n/p_1}-1, a^{n/p_2}-1,\dots, a^{n/p_m}-1)$

Question

Let $n=p_1^{\alpha_1}p_2^{\alpha_2}\dots p_m^{\alpha_m}$, it is possible to simplify $$L=\text{lcm}(a^{n/p_1}-1, a^{n/p_2}-1,\dots, a^{n/p_m}-1)$$ where $a\in\mathbb{N}$? It would also be fine to assume that $a$ is prime if that aids the progress. My thought was to use cyclotomic polynomials and maybe mess around with some identities of them. Clearly $$a^{n/\text{rad}(n)}-1\mid a^{n/p}-1$$ so we can pull this term out of the $\text{lcm}$ if desired, and if $m=2$ then using $$\gcd(a^{n/p_1}-1,a^{n/p_2}-1)=a^{n/(p_1p_2)}-1$$ we can get the result, but I cannot generalize. This simply came up in some stuff I was working on so it may not have a nice solution.

It would also help to find some $M$ such that $L\mid M$. Using one of the above observations one such $M$ could be $$M=\frac{\prod_p (a^{n/p}-1)}{(a^{n/\text{rad}(n)}-1)^{m-1}}$$ which is better than just trivially multiplying all the arguments together. Also as Thomas points out, $$M=\frac{a^n-1}{\Phi_n(a)}$$ works too.