Let $\alpha \in (0,1)$ and let $p_j$ be the sequence of primes and let $q_n:=\prod_{j\leq n}p_j$. Prove that $$ \sum_{p|q_n} (\log p)^\alpha \sim n(\log n)^\alpha \qquad n\to\infty $$
I am stuck here. Most what I proved is that $\sum_{p|q} (\log p)^\alpha \leq (\omega(q))^{1-\alpha} (\log q)^\alpha$, and then $\omega(q_n) = n$, but I am not sure if I can use this somehow ($\omega(n) = $ no. of distinct prime divisors of n). Any help?
Let $x:=p_n$. Then the sum under question equals $\sum_{p\leq x}(\log p)^\alpha$ which upon partial summation is not different to $$\pi(x)(\log x)^\alpha-\alpha\int_1^x\pi(x)\frac{(\log x)^{\alpha-1}}{x}\mathrm{d}x,$$ where $\pi(x)$ denotes the number of primes $p\leq x$. The prime number theorem in the form $\pi(x)=\frac{x}{\log x}(1+o(1))$ shows that the previous quantity is $$(1-\alpha) x(\log x)^{\alpha -1}(1+o(1)).$$ This can be proved either by using the $\epsilon-\delta$ definition of the limit of a sequence or any weak quantitative form of the prime number theorem. In order to express the final answer in terms of $n$ we observe that $$x=p_n=n(\log n)(1+o(1))\Rightarrow \log x \sim \log n$$ so that $$ (1-\alpha) x(\log x)^{\alpha -1}=(1-\alpha) n (\log n)^{\alpha}.$$ Is it possible that the leading factor $1-\alpha$ is missing from your question ?