Where is the error in my calculation for a simple bosonic system?

$ newcommand{ket}[1]{left|#1rightrangle}
$I was doing some calculations and I came across something that confused me and maybe you can explain to me where my reasoning is wrong. Let’s say I have some bosonic systems described by the tensor product of four independent states:
$$ ket{psi} = ket{n_1} otimes ket{n_2} otimes ket{n_3} otimes ket{n_3} tag{1}label{1}$$
The vacuum state is of course
$$ ket{varnothing} = ket{0} otimes ket{0} otimes ket{0} otimes ket{0} tag{2}label{2}$$
Let’s define some composite operator
$$ hat{A}=sum_i hat{a}_i tag{3}label{3}$$
where $hat{a}_i$ is the annihilation operator for the $i$-th state of the tensor product. The action of the adjunct of the operator $(ref{3})$ onto the vacuum state $(ref{2})$ is
$$ begin{align}
hat{A}^daggerket{varnothing}&= ket{1} otimes ket{0} otimes ket{0} otimes ket{0}
&quad+ket{0} otimes ket{1} otimes ket{0} otimes ket{0}
&quad+ket{0} otimes ket{0} otimes ket{1} otimes ket{0}
&quad+ket{0} otimes ket{0} otimes ket{0} otimes ket{1} tag{4a}label{4a}
&=ket{psi} otimes ket{psi} otimes ket{psi} otimes ket{psi} tag{4b}label{4b}
end{align}$$
where
$$ ket{psi} = ket{1} + 3ket{0} tag{5}$$
Now, repeating this identical for the operator $(ref{3})$ as written results in
$$ begin{align}
hat{A}ket{varnothing}&= 0 otimes ket{0} otimes ket{0} otimes ket{0}
&quad+ket{0} otimes 0 otimes ket{0} otimes ket{0}
&quad+ket{0} otimes ket{0} otimes 0 otimes ket{0}
&quad+ket{0} otimes ket{0} otimes ket{0} otimes 0 tag{6a}label{6a}
&=ket{psi} otimes ket{psi} otimes ket{psi} otimes ket{psi} tag{6b}label{6b}
end{align}$$
where
$$ ket{psi} = 3ket{0} tag{7}$$
which is obviously not correct, since the result should be the zero-vector. Where is my mistake? Is the step from $(ref{4a})$ to $(ref{4b})$ actually allowed? Clearly, for $(ref{6a})$ to $(ref{6b})$ this cannot be the case.