= Solution
The unit-normalized <Fock vacuum> is characterized by
$$
a_{\mathbf p}|0\rangle=0\quad\hbox{for every }\mathbf p,
\qquad \langle0|0\rangle=1.
$$
Choose the <relativistic normalization of a one-particle state> from the outset:
$$
|\mathbf p\rangle=\sqrt{2E_{\mathbf p}}\,a^\dagger_{\mathbf p}|0\rangle,
\qquad
|\mathbf p_1,\ldots,\mathbf p_n\rangle
=\prod_{r=1}^n\bigl(\sqrt{2E_{\mathbf p_r}}\,a^\dagger_{\mathbf p_r}\bigr)|0\rangle.
$$
Since any two <creation operators> commute, any <permutation> of their order gives exactly the same <state vector>. Thus these <multiparticle quantum states> are symmetric under exchange and describe <bosons>, without a minus sign or any additional exchange phase.
The <Hilbert space> is the <bosonic Fock space>, the direct sum of the symmetric $n$-particle <Hilbert spaces>, including the zero-particle sector. Sharp-<momentum> kets are distributionally normalized; square-integrable <wave packets> provide ordinary <Hilbert space> vectors. For repeated occupation of a discrete mode, a unit-normalized <number state> uses $(a^\dagger)^n|0\rangle/\sqrt{n!}$. That factorial does not belong in the above labelled continuum-ket convention.
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