In a Clifford algebra over an $n$-dimensional vector space $V$ with a non-degenerate quadratic/bilinear form, can any product of vectors $a_1,a_2,\cdots,a_m\in V$ be written as
$$a_1a_2\cdots a_m=a_1'a_2'\cdots a_k'$$
for some $k\leq n$ and $a_1',a_2',\cdots,a_k'\in V$?
Due to the grading on the algebra, necessarily $k\equiv m\bmod2$ (unless the product vanishes). It suffices to prove that a product of $n+1$ vectors can be reduced to a product of $n-1$ vectors.
If the vectors are all invertible ($a^{-1}=\frac{a}{a\cdot a}$), then they represent reflections ($v\mapsto-ava^{-1}=v-2\frac{v\cdot a}{a\cdot a}a$), and the product represents an isometry of $V$. According to the Cartan-Dieudonne theorem, any isometry is the composition of at most $n$ reflections, so we get the desired result. (I could make the correspondence between multivectors and isometries more explicit, but that's beside the point of this question.)
So, suppose some of the vectors are null ($a\cdot a=0$), and thus don't represent isometries.
Non-degeneracy means, for $a\neq0$, there's some vector $v\in V$ such that $a\cdot v\neq0$. To see that this condition is necessary, consider $V=\mathbb R^3$ with the degenerate quadratic form $v=(v_1e_1+v_2e_2+v_3e_3)\mapsto v\cdot v=v_1^2-v_2^2+0v_3^2$. (This space may be denoted $\mathbb R^{1,1,1}$.) The product of four unit vectors
$$e_1(e_1+e_3)e_2(-e_2+e_3)=1+(e_1+e_2)e_3$$
cannot be written as a product of two vectors; their wedge product would have to be $(e_1+e_2)e_3$, so they'd be in the span of the null vectors $e_1+e_2$ and $e_3$, but then their dot product would be $0$, not $1$ as required.
I've proven this for $n=2,3,4$ ($n=1$ requires a scalar in front of the empty product). See the Degenerate Cartan-Dieudonne chat. I also thought through a proof for $n=5$, but I didn't write it down, and it's a bit complicated. Anyway, the methods I used won't work for $n\geq6$. I relied on the fact that the product of vectors, $P=a_1a_2\cdots a_{n+1}$, has only grades $\equiv n+1\bmod2$, and has $P\tilde P$ and $\tilde PP$ being scalars. These latter properties alone imply that $P$ is a product of vectors, only in low dimensions. Over $\mathbb R^6$, the odd multivector $P=e_1e_2e_3+e_4e_5e_6$ has $P\tilde P=\tilde PP=2$, but $P$ is not a product of vectors, because $Pe_1\tilde P$ is not a vector (it's a $5$-vector).
