As Mikhail Tikhomirov commented, the denominator should have $\log$ of $p_j$ instead of $p_j$ itself.

For a reference see Theorem 5.3, and more generally, section 2 (`The geometric method') of Chapter 5 of part III of G. Tenenbaum's book ``Introduction to analytic and probabilistic number theory''. This section is contained in pages 516-517.

If we let
$$N_k(x):=\left| \left\{ (\nu_1,\ldots,\nu_k) \in \mathbb{N}^k: \sum_{1 \le j \le k} \nu_j a_j \le x\right\} \right|$$
for given positive $a_j$, the theorem states that
$$\frac{x^k}{k!}\prod_{1\le j\le k}\frac{1}{a_j} \le N_k(x)\le \frac{(x+\sum_{1 \le j \le k}a_j)^k}{k!}\prod_{1\le j\le k}\frac{1}{a_j} .$$
Now apply this with $a_j = \log p_j$. This answers your question, because you count solutions to $\prod_{1 \le j \le k} p_j^{a_j} \le x$, which (upon taking natural logarithms) we see is a quantity equal to $N_k(x)$.
This is directly related to V. Ennola's work on $y$-friable (i.e. smooth) numbers, which is mentioned in the aforementioned section. It could be that your result is in Ennola's paper ``On numbers with small prime divisors'' (Ann. Acad. Sci. Fenn., Ser. A I 440, 16 p. (1969)) but I am not able to get my hands on it.