Geometric algebra: is a $k$-vector simple iff its square is a scalar? In Geometric Algebra, a multivector $u$ is simple if it is of the form $u = v_1\wedge \cdots\wedge v_k$ for some vectors $v_i$ (a.k.a., $u$ is a $k$-blade).
Clearly, a multivector can only be simple if it a $k$-vector, that is, if it is not a mixed multivector.
Is there an easy test of whether a given $k$-vector is simple?
‘Easy’ meaning something that could be implemented algorithmically.
(This is perhaps related to the problem to diagnosing whether a differential $p$-form is simple.)

My conjecture is that a $k$-vector $u$ is simple iff $u^2$ is a scalar, where $u^2 = uu$ is the geometric product of $u$ with itself.
I think I have shown the $\implies$ direction (see below), and have not been able to find a counterexample to the $\Longleftarrow$ direction: that is, an example of a $k$-vector whose geometric-square is a scalar but which is not simple.

Proof that a simple $k$-vector has scalar geometric-square:
Let $u = v_1\wedge\cdots\wedge v_k$ be any simple $k$-vector. We may write
\begin{align}
u &= v_1\wedge v_2\wedge \cdots v_k
\\ &= \underbrace{\left(v_1 - \operatorname{proj}_{v_2}(v_1)\right)}_{\bar v_2}\wedge v_2\wedge\cdots\wedge v_k
,\end{align}
where $\operatorname{proj}_{v_2}(v_1) = \frac{v_1\cdot v_2}{v_2\cdot v_2}v_2$ since $\operatorname{proj}_{v_2}(v_1)\wedge v_2 = 0$. Notice that $\bar v_1  \cdot v_2 = 0$ by construction, and therefore $\bar v_1 \wedge v_2 = \bar v_1 v_2$ (since $ab = a\cdot b + a\wedge b$ for $1$-vectors). We proceed inductively, using associativity:
\begin{align}
u &= \bar v_1 (v_2\wedge v_3) \cdots \wedge v_k
\\ &= \bar v_1 \left(v_2 - \operatorname{proj}_{v_3}(v_2)\right)\wedge v_3 \wedge \cdots \wedge v_k
\\ &= \bar v_1 \bar v_2 v_3 \wedge \cdots \wedge v_k
\\&\ \;\vdots
\\ &= \bar v_1\bar v_2 \cdots \bar v_k
\end{align}
Finally, we normalise $e_i := \bar v_i/\|\bar v_i\|$ and introduce an appropriate scaling factor $\lambda = \|\bar v_1\|\cdots\|\bar v_k\|$.
Notice that $\{e_i\}$ is orthonormal.
Thus, $u$ is a $k$-blade, whose square is trivially a scalar:
$$ uu = \lambda^2 e_{i_1}\cdots e_{i_k} e_{i_1}\cdots e_{i_k} = \pm \lambda^2 e_{i_1}^2\cdots e_{i_k}^2 = \pm \lambda^2 \|e_{i_1}\|^2\cdots \|e_{i_k}\|^2.$$
(In the middle equality, a factor of $\pm1$ is generated from commuting the $e_i$’s which satisfy $e_ie_j = -e_je_i$ if $i\ne j$.)

 A: Actually even your forward direction is incorrect: it is not true that the square of what you call a simple $k$-vector is a scalar (or a pseudo-scalar, or a central element). The mistake in your proof is that you cannot write $u=\lambda e_{i_1}\cdots e_{i_k}$ in general.
As a counter-example (assuming $(e_i)$ is an orthonormal basis): $u=e_1\cdot (e_1+e_2)=e_1e_2+1$ (so $v_1=e_1$ and $v_2=e_1+e_2$) gives
$$u^2=(e_1e_2)^2+2e_1e_2+1=2e_1e_2$$
which is not central or anything if the dimension of the quadratic space is at least $3$. 
A: Clarification: In this context, “$u$ is simple” means “$u$ is a $k$-blade” means “$u = \vec u_1\wedge \cdots \wedge \vec u_k$ for some vectors $\vec u_i$”.
Answer: No. $u$ is simple $\Longrightarrow$ $u^2$ is scalar $\,\,\not\!\!\Longrightarrow$ $u$ is simple
Proof:
If $u$ is simple, then there exists an orthonormal basis for which $u = λe_1\cdots e_k$; then $u^2 = (-1)^{\frac{k(k-1)}2}\tilde{u}u = (-1)^{\frac{k(k-1)}2}λ^2 e_1^2 \cdots e_k^2$ is a scalar.
A counter-example to the reverse implication is $u = e_1e_2e_3 + e_4e_5e_6$.
This is a $3$-vector which is not simple, since it cannot be factored into $u = \vec u_1 \wedge \vec u_2 \wedge \vec u_3$.
However, its square is a scalar: $u^2 = -e_1^2e_2^2e_3^2 - e_4^2e_5^2e_6^2$ which is $-2$ assuming a Euclidean signature $e_i^2 = 1$.
