There is a paper I'm reading On Miki's Identity for Bernoulli Numbers that states the following without proof.
$$\frac{1}{(1-x)(1-2x)...(1-mx)}=\exp\left[\sum_{k=1}^\infty\left(1^k+2^k+...+m^k\right)\frac{x^k}{k!}\right]$$
I can split the product up and rewrite the left as
$$\left(\sum_{k=0}^\infty{x^k}\right)\left(\sum_{k=0}^\infty{2^kx^k}\right)...\left(\sum_{k=0}^\infty{m^kx^k}\right)=\left(\sum_{k=0}^\infty{k!\frac{x^k}{k!}}\right)\left(\sum_{k=0}^\infty{2^kk!\frac{x^k}{k!}}\right)...\left(\sum_{k=0}^\infty{m^kk!\frac{x^k}{k!}}\right)$$
On the right, I can rewrite as
$$\exp{\left[\exp{(x)}-1\right]}\exp{\left[\exp{(2x)}-1\right]}...\exp{\left[\exp{(mx)}-1\right]}$$
And I don't feel like I'm making headway.
Hint:
Prove the correct problem by induction:
You essentially need this:
$$\frac1{1-mx}=\exp\left[\sum_{n=1}^\infty m^kx^k/k\right]$$
And as a hint, you may want to use the Taylor expansion of the logarithm.