Let $H$ be a Hilbert space and let $V \subseteq H$ be a dense subspace. The exterior algebra \begin{equation} \oplus_{n=0}^{\infty} \wedge^{n}H \end{equation} can be equipped with a canonical inner product that turns it into a pre-Hilbert space, let us write $\Lambda H$ for its Hilbert completion.
Is it true that $\Lambda V = \Lambda H$?
The following seems like a reasonable proof strategy to me.
We could try to show that \begin{equation} \left( \oplus_{n=0}^{\infty} \wedge^{n}V \right)^{\perp} = \{0 \}. \end{equation} This could be done by showing first that \begin{equation} \left( \oplus_{n=0}^{\infty} \wedge^{n}V \right)^{\perp} = \oplus_{n=0}^{\infty} \left(\wedge^{n}V \right)^{\perp}, \end{equation} where on the right hand side we have that the orthocomplement of $\wedge^{n}V$ is taken in $\wedge^{n}H$. The second step would be to show that, for each $n$, we have that $\wedge^{n}V$ is dense in $\wedge^{n}H$, which would imply that \begin{equation} \wedge^{n}(V)^{\perp} = \{ 0 \}. \end{equation}
I guess this second step is what I'm not too sure about.