Is the relationship between these two sequences, identical but for signs, trivial? Sums of divisors of product of first $n$ primes has same sequence as products of totients of first $n$ primes, but with alternating signs. OEIS A005867
Mathematica code shown:

Table[Sum[
    MoebiusMu[d] d, {d, Divisors[Product[Prime[j], {j, 1, n}]]}], {n, 1,
     8}]
  Table[Product[EulerPhi[Prime[j]], {j, 1, n}], {n, 1, 8}]

{-1, 2, -8, 48, -480, 5760, -92160, 1658880}
{1, 2, 8, 48, 480, 5760, 92160, 1658880}

Edit an alternate for the first statement:
$$\sum_{n=1}\ \sum_{(d_{1},\dots,d_{2^{n}})\ |\ \text{lcm}(p_{1},\dots,p_{n})}\mu(d_{m})d_{m}$$
Is this a trivial relationship?

Yes, this is trivial — at least for some definition of trivial. The sum is literally a term-by-term expansion of the product of $(p_i-1)$; it iterates all the members of the product, with the sign of a given term $\Pi_{i\in S}p_i$ determined by the parity of the number of distinct primes $|S|$ in the individual term - in other words, by $\mu(\Pi_{i\in S}p_i)$.
