Asymptotic for the probability that a number has $k$ prime factors less than $Q$ 
If we let $\omega_Q(n)$ denote the number of distinct prime factors of $n$ less than a bound $Q$, then what asymptotic formulas exist for $\Pr_{n\in\mathbb{N}}[\omega_Q(n)=k]$ as $Q\to\infty$ if $k$ remains fixed (or perhaps very small with respect to n)?

I am asking this question since my study led me to want to bound the quantity
$$\mathbf{E}_{n\in\mathbb{N}}\left[\frac{2^{\omega_Q(n)}}{\sqrt{\omega_Q(n)}}\right]$$
as $Q\to\infty$. Since
$$\mathbf{E}_{n\in\mathbb{N}}\left[\frac{2^{\omega_Q(n)}}{\sqrt{\omega_Q(n)}}\right]=\sum_{n=1}^{\pi(Q)}\left(\Pr_{n\in\mathbb{N}}[\omega_Q(n)=k]\right)\left(\frac{2^{\omega_Q(n)}}{\sqrt{\omega_Q(n)}}\right)$$
and
$$\sum_{n=1}^{\pi(Q)}\Pr_{n\in\mathbb{N}}[\omega_Q(n)=k]2^{\omega_Q(n)}\sim_{Q\to\infty} c\log(Q)$$
is well understood, good (upper) bounds on $\Pr_{n\in\mathbb{N}}[\omega_Q(n)=k]$ could help me in my effort.
For small values of $k$ computations can be done directly, like
$$\Pr_{n\in\mathbb{N}}[\omega_Q(n)=0]\sim\frac{c}{\log(Q)}$$
and
$$\Pr_{n\in\mathbb{N}}[\omega_Q(n)=1]\sim c\frac{\log(\log(Q))}{\log(Q)}$$
The main approach I have been using is noting that $\Pr_{n\in\mathbb{N}}[\omega_Q(n)=k]$ is exactly the coefficient of $x^k$ in the polynomial
$$\prod_{p<Q}\left(\frac{x}{p}+1-\frac{1}{p}\right)$$
Asymptotics of this full polynomial are easy to come by, for instance as $Q\to\infty$ we have that
$$\prod_{p<Q}\left(\frac{x}{p}+1-\frac{1}{p}\right)\sim c \log^{x-1}(Q)$$
Heuristically this would suggest that
\begin{align*}
\Pr_{n\in\mathbb{N}}[\omega_Q(n)=k]&=\frac{1}{k!}\left.\frac{d^k}{dx^k}\prod_{p<Q}\left(\frac{x}{p}+1-\frac{1}{p}\right)\right|_{x=0}\\
&\sim \frac{c}{k!}\left.\frac{d^k}{dx^k}\log^{x-1}(Q)\right|_{x=0}\\
&=\frac{c}{k!}\frac{\log^k(\log(Q))}{\log(Q)}
\end{align*}
This argument is however by no means rigorous so I would appreciate true asymptotics.
 A: As pointed out in the question, we have that
$$\prod_{p<Q}\left(\frac{x-1}{p}+1\right)=\sum_{k=0}^{\pi(Q)}\Pr_{n\in\mathbb{N}}[\omega_Q(n)=k]x^k$$
which can be derived by showing that on both the RHS and the LHS the coefficient of $x^k$ is equal to
$$\sum_{\substack{S\subseteq \{p<Q\} \\ |S|=k}} \left(\prod_{p\in S}\frac{1}{p}\right)\left(\prod_{p\not\in S}\left(1-\frac{1}{p}\right) \right)$$
Treating the LHS with simple manipulation we get that
\begin{align*}
\prod_{p<Q}\left(\frac{x}{p}+1-\frac{1}{p}\right)&=\exp\left(\log\prod_{p<Q}\left(\frac{x}{p}+1-\frac{1}{p}\right)\right)\\
&=\exp\left(\sum_{p<Q}\log\left(\frac{x-1}{p}+1\right)\right)\\
&=\exp\left(\sum_{p<Q}\frac{x-1}{p}+\sum_{p<Q}\left(\log\left(\frac{x-1}{p}+1\right)-\frac{x-1}{p}\right)\right)\tag{1}
\end{align*}
We now note that
$$\sum_{p<Q}\left(\log\left(\frac{x-1}{p}+1\right)-\frac{x-1}{p}\right)=\sum_{p}\left(\log\left(\frac{x-1}{p}+1\right)-\frac{x-1}{p}\right)-\sum_{p\geq Q}\left(\log\left(\frac{x-1}{p}+1\right)-\frac{x-1}{p}\right)$$
and thus we can set
$$f_1(x)=\sum_{p}\left(\log\left(\frac{x-1}{p}+1\right)-\frac{x-1}{p}\right)$$
and
$$g(x)=\sum_{p\geq Q}\left(\log\left(\frac{x-1}{p}+1\right)-\frac{x-1}{p}\right)$$
Morally, we can think of $g(x)$ as the "error" as $Q\to\infty$ which we must show is inconsequential. We thus get that
\begin{equation}
\sum_{p<Q}\left(\log\left(\frac{x-1}{p}+1\right)-\frac{x-1}{p}\right)=f_1(x)-g(x)\tag{2}
\end{equation}
By Merten's theorem, we have that
\begin{align*}
\sum_{p<Q}\frac{x-1}{p}&=(x-1)\left(\log(\log(Q))+M+\epsilon_Q\right)\\
&=(x-1)\log(\log(Q))+Mx+\epsilon_Q x-M-\epsilon_Q\tag{3}
\end{align*}
where $\epsilon_Q\to 0$ and $M$ is the Meissel-Mertens constant. Substituting (2) and (3) into (1) we get that
\begin{align*}
\prod_{p<Q}\left(\frac{x}{p}+1-\frac{1}{p}\right)&=\exp\left((x-1)\log(\log(Q))+Mx+\epsilon_Q x-M-\epsilon_Q+f_1(x)-g(x)\right)\\
&=e^{-M-\epsilon_Q}\log^{x-1}(Q)e^{Mx}e^{f_1(x)}e^{\epsilon_Qx}e^{-g(x)}
\end{align*}
For simplicity's sake, we now define
$$f_2(x)=e^{Mx}e^{f_1(x)}$$
and thus
\begin{equation}
\prod_{p<Q}\left(\frac{x}{p}+1-\frac{1}{p}\right)=e^{-M-\epsilon_Q}\log^{x-1}(Q)f_2(x)e^{\epsilon_Qx}e^{-g(x)}\tag{4}
\end{equation}
Taking the derivative $k$ times yields
$$\frac{d^k}{dx^k}e^{-M-\epsilon_Q}\log^{x-1}(Q)f_2(x)e^{\epsilon_Qx}e^{-g(x)}$$
As $Q\to\infty$, the only term that will matter in a product rule decomposition of this equation is the one that grows the fastest. It is easy to show that
$$g^{(n)}(x)=O\left(\frac{1}{x}\right)$$
for any order derivative $(n)$, and so the fastest growing term is the one where $\log^{(x-1)}(Q)$ is differentiated the full $k$ times. Since there are finitely many terms the others are inconsequential in terms of growth and so
\begin{align*}
\frac{d^k}{dx^k}e^{-M-\epsilon_Q}\log^{x-1}(Q)f_2(x)e^{\epsilon_Qx}e^{-g(x)}&\sim_{Q\to\infty}e^{-M-\epsilon_Q}f_2(x)e^{\epsilon_Qx}e^{-g(x)}\frac{d^k}{dx^k}\log^{x-1}(Q)\\
&=e^{-M-\epsilon_Q}f_2(x)e^{\epsilon_Qx}e^{-g(x)}\log^k(\log(Q))\log^{x-1}(Q)
\end{align*}
evaluating at $x=0$ and substituting into (4) yields that
\begin{equation}
\left.\frac{d^k}{dx^k}\prod_{p<Q}\left(\frac{x}{p}+1-\frac{1}{p}\right)\right|_{x=0}\sim e^{-M-\epsilon_Q}f_2(0)e^{-g(0)}\frac{\log^k(\log(Q))}{\log(Q)}
\end{equation}
As $Q\to\infty$ we have that $g(0)\to0$ and $f_2(0)=e^{M-\gamma}$ and so
\begin{equation}
\left.\frac{d^k}{dx^k}\prod_{p<Q}\left(\frac{x}{p}+1-\frac{1}{p}\right)\right|_{x=0}\sim e^{-\gamma}\frac{\log^k(\log(Q))}{\log(Q)}\tag{5}
\end{equation}
We also see that
\begin{align*}
\left.\frac{d^k}{dx^k}\prod_{p<Q}\left(\frac{x}{p}+1-\frac{1}{p}\right)\right|_{x=0}&=\left.\frac{d^k}{dx^k}\sum_{j=0}^{\pi(Q)} \mathrm{Pr}_{n\in\mathbb{N}}[\omega_Q(n)=j]x^j\right|_{x=0}\\
&=k!\mathrm{Pr}_{n\in\mathbb{N}}[\omega_Q(n)=k]
\end{align*}
and thus we conclude from (5) that
$$\mathrm{Pr}_{n\in\mathbb{N}}[\omega_Q(n)=k]\sim e^{-\gamma}\frac{\log^k(\log(Q))}{\log(Q) k!}$$
which is the desired result
