My solution in the context of $n$-dim cube is the following. Obviously, that there are $2^n$ vertices in $n$-dim cube. Then at each vertex choose $m$ $1$-dim edges which are part of $m$-dim cube. You will obtain $2^n C_{n}^{m}$. But you should divide this number by the number of how many times the same $m$ -dim cube was added in total sum which is $2^m$. Voilà, $Q_{n}^{m} = 2^{n-m}C_{n}^{m}.$

Thank you for this solution. I obtained the same result before. Now I can reveal the motivation. I will add it in the question description,

Good catch! I fixed this typo. I will try with generating function.

@js21, I agree. What an obvious fact that I did not note. Thank you.

@js21, Of course $\beth_k(a)$ is define for only $a\geq k$. I will add your remark to the description. P.S, I have tested $\beth_2(n)$ for multiplicity for $n$ up to 100.