Prove that this language is NP-Hard Given
$$\mathrm{\#3SAT} = \{ (w, y) \mid  w\text{ is a $\mathrm{3SAT}$ instance with at least $y$ satisfying assignments}\}\,,$$
prove that $\mathrm{\#3SAT}$ is NP-Hard. 

I am currently stuck with this one. 

Basically $\mathrm{\#3SAT}$ is the counting version of $\mathrm{3SAT}$ and to me it is seems clear that establishing membership in the language is more complex than establish the classic $\mathrm{3SAT}$ membership.

I'm trying to reason through the prover-verifier paradigm: let's say you have a $\mathrm{3SAT}$ instance, now suppose the certificate for that instance is $n$&nbsp;bits. It is obvious that the certificate for the same instance but translated into $\mathrm{\#3SAT}$ with let's say $y=10$ would be at least $10n$ bits.  So if the first can be examined in time $t$ then the second will take at least $10t$ and so on. However I am not satisfied with this reasoning, it seems  incomplete and superficial. I would greatly appreciate your advice on how to proceed with this proof. Thanks in advance.
It seems that you're trying to prove that $\#\mathrm{3SAT}$ is in $\mathrm{NP}$. Since $\#\mathrm{3SAT}$ is $\mathrm{\#P}$-complete, and $\mathrm{\#P}$ seems to be harder than anything in the polynomial hierarchy, it's very unlikely that $\#\mathrm{3SAT}\in\mathrm{NP}$.

The error in your attempted proof is that your certificate doesn't have polynomial length. A satisfying assignment has length $\Theta(|x|)$ (the formula can't have more variables than its length, but it could be of the form $(x_1\lor x_2\lor x_3) \land (x_4\lor x_5\lor x_6) \land\dots$). The number&nbsp;$y$ could have any value between zero and $2^{|y|}-1$, so your certificate has length about $|x|2^{|y|}$, which is exponential in the input size.

However, the question asks you to prove that it's $\mathrm{NP}$-hard, not that it's $\mathrm{NP}$-complete. You just need to prove that there's a polynomial-time reduction from some $\mathrm{NP}$-complete problem to $\#\mathrm{3SAT}$.