Let $V$ be a $m$ dimensional vector space over a field $\mathbb{F}$ and $1\leq l<m.$ Then consider the wedge product $\bigwedge^lV$ and $\bigwedge^{m-l}V.$ Fix a basis $\{e_1,\ldots,e_m\}$ of $V$ and cosider $I(l,m)=\{(\alpha_1,\ldots,\alpha_l)\in \mathbb{Z}^l:1\leq \alpha_1 <\alpha_2\cdots<\alpha_l \leq m\}.$ We know that $\{e_{\alpha}: \alpha \in I(l,m)\}$ where $e_{\alpha}=e_{\alpha_1} \wedge \cdots \wedge e_{\alpha_l}$ forms a basis for $\bigwedge^lV$. Now for $\alpha \in I(l,m)$ let $\alpha ^c=(\alpha_{1}^c,\ldots, \alpha_{m-l}^c)$ denote the unique element of $I(m-l,m)$ such that $\{\alpha_1,\ldots, \alpha_l\}\cup\{\alpha_{1}^c,\ldots, \alpha_{m-l}^c\}=\{1,\ldots,m\}.$ So $\{e_{\alpha}^c:\alpha \in I(l,m)\}$ gives a basis for $\bigwedge^{m-l}V.$ Now we define well known Hodge star operator $h: \bigwedge^lV \to \bigwedge^{m-l}V$ by defining on the basis above as following. Define $h(e_{\alpha})=(-1)^{\alpha_1 + \cdots+\alpha_l+l(l+1)/2}e_{\alpha^c}$ for $\alpha \in I(l,m).$ We also call an non zero element $w \in \bigwedge^lV$ decomposable if $w=x_1\wedge x_2 \wedge \cdots x_l $ for some $\{x_1, \ldots, x_l\}\subset V.$
Now I want to show that image of a decomposable element under h is decomposable and conversely.
I need some help to prove that.
It suffices to prove only one direction as $h\big(h(w)\big)=(-1)^{l(m-l)}\,w$ for every $w\in \bigwedge^l V$. That is, we shall only verify that, if $w\in \bigwedge^l\,V$ is decomposable, then $h(w)\in \bigwedge^{m-l}\,V$ is also decomposable.
Equip $V$ with the nondegenerate bilinear form $\langle\_,\_\rangle$ defined by $$\left\langle\sum_{i=1}^m\,v^ie_i,\sum_{i=1}^m\,w^ie_i\right\rangle:=\sum_{i=1}^m\,v^iw^i$$ for every $v^1,v^2,\ldots,v^m,w^1,w^2,\ldots,w^m\in \mathbb{F}$. Then, for each subset $S$ of $V$, we write $S^\perp$ for the orthogonal complement of $S$ with respect to $\langle\_,\_\rangle$, i.e., $S$ is the subspace of $V$ consisting of all $v\in V$ such that $\langle s,v\rangle =0 $ for all $s\in S$.
Suppose now that $w=x_1\wedge x_2\wedge \cdots \wedge x_l$ for some $S:=\left\{x_1,x_2,\ldots,x_l\right\}\subseteq V$. If $w=0$, then there is nothing to prove. We assume from now on that $w \neq 0$. Hence, $S$ consists of $l$ linearly independent elements of $V$. Therefore, $S^\perp$ is an $(m-l)$-dimensional subspace of $V$. We claim that $h(w) \wedge v=0$ for every $v\in S^\perp$. From this claim, it follows that $h(w)=y_1 \wedge y_2 \wedge \cdots \wedge y_{m-l}$, where $\left\{y_1,y_2,\ldots,y_{m-l}\right\}$ is a basis of $S^\perp$.
For each $j=1,2,\ldots,l$, write $x_j$ as $\sum_{i=1}^m\,x_j^ie_i$. Thus, $$w=\sum_{\alpha\in I(l,m)}\,w^\alpha e_\alpha\,,$$ where, for every $\alpha:=(i_1,i_2,\ldots,i_l)\in I(l,m)$, $$w^\alpha:=\det\left(\begin{bmatrix} x_1^{i_1}&x_2^{i_1}&\cdots&x_l^{i_1}\\ x_1^{i_2}&x_2^{i_2}&\cdots&x_l^{i_2}\\ \vdots&\vdots&\ddots&\vdots\\ x_1^{i_l}&x_2^{i_l}&\cdots&x_l^{i_l} \end{bmatrix}\right)\,.$$ Then, $$h(w)=(-1)^{\frac{l(l+1)}{2}}\,\sum_{\alpha \in I(l,m)}\,(-1)^{s(\alpha)}\,w^\alpha e_{\alpha^\complement}\,,$$ where $s(\alpha)=i_1+i_2+\ldots+i_l$ for $\alpha:=(i_1,i_2,\ldots,i_l)\in I(l,m)$.
By abuse of notation, we say that $i\in \alpha :=\left(i_1,i_2,\ldots,i_l\right)\in I(l,m)$ if $i = i_\mu$ for some $\mu=1,2,\ldots,l$. Now, for $v\in S^\perp$, write $v=\sum_{i=1}^m\,v^ie_i$. Define, for each $\beta:=(j_1,j_2,\ldots,j_{l-1}) \in I(l-1,m)$, $$f(\beta,v):=\sum_{i\in \beta^\complement}\,v^i\,\det\left(\begin{bmatrix} x_1^{j_1}&x_1^{j_2}&\cdots&x_1^{j_{l-1}}& x_1^i \\ x_2^{j_1}&x_2^{j_2}&\cdots&x_2^{j_{l-1}}& x_2^i \\ \vdots&\vdots&\ddots&\vdots&\vdots \\ x_{l-1}^{j_1}&x_{l-1}^{j_2}&\cdots&x_{l-1}^{j_{l-1}}&x_{l-1}^i \\ x_l^{j_1}&x_l^{j_2}&\cdots&x_l^{j_{l-1}}&x_l^i \end{bmatrix} \right)\,,$$ which is equal to $$\det\left( \begin{bmatrix} x_1^{j_1}&x_1^{j_2}&\cdots&x_1^{j_{l-1}}&\langle v,x_1\rangle \\ x_2^{j_1}&x_2^{j_2}&\cdots&x_2^{j_{l-1}}&\langle v,x_2\rangle \\ \vdots&\vdots&\ddots&\vdots&\vdots \\ x_{l-1}^{j_1}&x_{l-1}^{j_2}&\cdots&x_{l-1}^{j_{l-1}}&\langle v,x_{l-1}\rangle \\ x_l^{j_1}&x_l^{j_2}&\cdots&x_l^{j_{l-1}}&\langle v,x_l\rangle \end{bmatrix} \right)=0\,.$$ For $\beta\in I(l-1,m)$ and $i\in \beta^\complement$, we write $\beta\cup i$ for the $l$-tuple obtained from $\beta$ by adding an entry $i$ so that the entries of the newly created $l$-tuple form an increasing sequence. For $\beta \in I(l-1,m)$ and $i\in\beta^\complement$, write $\nu(i,\beta)$ for the number of entries in $\beta$ that are greater than $i$. Then, $$f(\beta,v)=\sum_{v\in \beta^\complement}\,(-1)^{\nu(i,\beta)}v^iw^{\beta\cup i}\,.$$
Observe that $w^\alpha e_{\alpha^\complement} \wedge v^ie_i=0$ if $i\in\alpha^\complement$, and $$w^\alpha e_{\alpha^\complement} \wedge v^i e_i =(-1)^{\epsilon(i,\alpha)}v^i w^\alpha e_{(\alpha\setminus i)^\complement}$$ if $i\in\alpha$, where $\alpha \setminus i$ is the $(l-1)$-tuple oobtained from $\alpha$ by removing $i$, and $\epsilon(i,\alpha)$ is the number of entries of $\alpha^\complement$ that are larger than $i$.
We now write $$h(w)\wedge v=(-1)^{\frac{l(l+1)}{2}}\,\sum_{\alpha \in I(l,m)}\,\sum_{i\in\alpha}\,(-1)^{s(\alpha)+\epsilon(i,\alpha)}v^iw^\alpha e_{(\alpha\setminus i)^\complement}\,.$$ Then, $$h(w)\wedge v = (-1)^{\frac{l(l+1)}{2}}\,\sum_{\beta \in I(l-1,m)}\,(-1)^{s(\beta)}e_{\beta^\complement}\,\sum_{i\in\beta^\complement}\,(-1)^{i+\epsilon(i,\beta\cup i)}v^iw^{\beta\cup i}\,.$$
To finish the proof, we note that $$i+\epsilon(i,\beta\cup i)+\nu(i,\beta)=m$$ for each $\beta \in I(l-1,m)$ and $i\in\beta^\complement$. This means $$h(w)\wedge v=(-1)^{\frac{l(l+1)}{2}+m}\,\sum_{\beta \in I(l-1,m)}\,(-1)^{s(\beta)}f(\beta,v)\,e_{\beta^\complement}=0\,.$$ The proof is now complete.
Examples
Let $\mathbb{F}:=\mathbb{F}_3$ and $V:=\mathbb{F}_3^3$ with standard basis vectors $e_1,e_2,e_3$. Thus, with $w:=e_1+e_2+e_3$, we have $S=\left\{e_1+e_2+e_3\right\}$, so that $S^\perp=\text{span}\left\{e_1+e_2+e_3,e_2+2e_3\right\}$ and $$h(w)=e_{(2,3)}+2e_{(1,3)}+e_{(1,2)}=(e_1+e_2+e_3)\wedge (e_2+2e_3)\,.$$
Let $\mathbb{F}:=\mathbb{F}_2$ and $V:=\mathbb{F}_2^4$ with standard basis vectors $e_1,e_2,e_3,e_4$. Then, with $$w:=(e_1+e_2)\wedge (e_3+e_4)=e_{(1,3)}+e_{(1,4)}+e_{(2,3)}+e_{(2,4)}\,,$$ we see that $S=\left\{e_1+e_2,e_3+e_4\right\}$ which spans $S^\perp$. Indeed, $$h(w)=w=(e_1+e_2)\wedge (e_3+e_4)\,.$$
Let $\mathbb{F}:=\mathbb{Q}$ and $V:=\mathbb{Q}^5$ with standard basis vectors $e_1,e_2,e_3,e_4,e_5$. For $$w:=(e_1-e_2)\wedge (e_3+e_4+e_5)=e_{(1,3)}+e_{(1,4)}+e_{(1,5)}-e_{(2,3)}-e_{(2,4)}-e_{(2,5)}\,,$$ we get $S=\left\{e_1-e_2,e_3+e_4+e_5\right\}$ and $S^\perp=\text{span}\left\{e_1+e_2,e_3-e_4,-e_3+e_5\right\}$. In addition, $$\begin{align}h(w)&=-e_{(2,4,5)}+e_{(2,3,5)}-e_{(2,3,4)}-e_{(1,4,5)}+e_{(1,3,5)}-e_{(1,3,4)} \\&=(e_1+e_2)\wedge (e_3-e_4)\wedge (-e_3+e_5)\,.\end{align}$$