I guess one can prove it by using $\hat{a}_{\phi_1}\dots\hat{a}_{\phi_N}\Psi = 0$, which means $\Psi$ is orthogonal to every Slater determinant, and if all Slater determinants is a dense subset of all $N$-electron wave functions we are done.

No. The correct statement is that for $\psi \in F(\mathfrak h)$ it holds that: If $a(f)\psi=0$ for all $f\in \mathfrak h$, then $\psi=\Omega$, where $\Omega$ is any state in the zero-particle sector of $F(\mathfrak h)$. In particular, no $N>0$ vector can obey $a(f)\psi=0$ for all $f\in \mathfrak h$.

Do you mean, "Is $|\Psi\rangle=0$?" or, "Is $|\Psi\rangle=|0\rangle$?"

In any case, what I want to add to my answer: Obviously the zero vector is a solution of $a(f)\psi=0$. The point is that it has a non-trivial (non-vansihing/zero) solution, namely multiples of the vacuum vector. However, if you from the very beginning restrict $\psi$ to be a $N>0$ particle state, then yes, the only solution is $\psi=0$.

@TobiasFünke I agree with you. I should add in my questions $N > 0$.

@Rasmus that seems like producing a contradiction, if you did that.

In the setup of your question, you state that $\Psi$ is an $N$-electron state with $N > 0$. The vector $\Psi = 0$ is not such a state, so it seems trivially true that $\Psi$ cannot be zero without even needing to consider your main equation.

@TobiasFünke The last sentence in your previous comment seems like a contradiction. The state $\psi = 0$ is not an $N \geq 0$ particle state - or indeed, any state at all.

@tparker, yes, states are non-zero. What should I call a state that is allowed to be zero? Is wave function better?

@tparker the "$N>0$" was edited in after the comment of TobiasFünke. Calling the zero element of the Hilbert-space a state or not seems like a rather linguistic discussion, how would you refer to it?

@Zaph I say that it's a vector in the Hilbert space that does not correspond to a physical state.

@tparker ??? I don't understand. It is a vector in the Fock space--I think you are, for whatever reason, now nitpicking with terminology used by physicists. As I said, it is a well-known theorem for both fermions and bosons that $a(f)\psi=0$ for $\psi\in F(\mathfrak h)$ and all $f\in \mathfrak h$ implies that $\psi=c\Omega$, where $c\in \mathbb C$, and in particular can be $0$.

@J.Delaney This will be my last comment here (I cannot state every mathematical detail in the comment section): Any $n$ particle vector $\psi_n\in H_n$ can be naturally identified with a corresponding element of the Fock space $\tilde \psi_n:=(0,0,\ldots, \psi_n,0,0,\ldots)\in F$. What I meant is that the only solution of $a(f)\psi=0$ for all $f$ is $\psi=c\Omega$. If $\psi=\tilde \psi_n$ for some $n\in\mathbb N_0$, this means that $n=0$ or $\psi_n=0$ if $n>0$, which in this case also implies $\tilde \psi_n=0\in F$.

@TobiasFünke I see what you mean, but this is not the definition of an "$n$-particle state", at least not as used by physicists: of course $\psi=0 \in H_n$ for every $n$, but to be called an "$n$-particle state" $\psi$ has to have a unique $n$ associated with it, i.e. it needs to be a non-zero vector in the eigen-subspace of the number operator corresponding to the eigenvalue $n$.

@J.Delaney I don't see what all the nitpicking/arguing about the terminology should lead to. The question of the OP is (now), IMHO, more than clear. You can define what you want, you can call things like you want... but given the limits of symbols in comments, I think my comments should be more or less clear; if read in good and not bad faith. Personally, I would call $0\in H_n$ a $n$-particle vector (but not $0=(0_0,0_1,0_2,0_3,\ldots)\in F$); but okay, you do not have to agree with. Still the theorem I've written (if rigorously formulated) holds true, which is the important thing.

@TobiasFünke I guess you can see by the votes that the question as stated doesn't make sense to a lot people. A more accurate way to state it, in my opinion, is to say that $\psi$ that satisfies the given condition is either the vacuum ($N=0$) or the zero vector.

I think I do agree with J. Delaney here. When talking about physical states we usually assume they have norm 1, or at least can be normalized by multiplication with some scalar. When an annihilation operator acting on such a state gives zero for all orbitals, that state must be empty, which is a contradiction to being an N-particle state. I think it would be very beneficial to change the formulation into “Vector in an N-particle-Hilbert space”, so people don’t assume it should lie on the sphere-surface.