If $K$ is a Kakeya set in $F_q^m$ then $K\times F_q^{n-m}$ is a Kakeya set in $F_q^n$? Prove or disaprove the following statment:

Let $q$ the power of a prime number $p$ and $n,m\in \mathbb{Z}^{+}$ such that $n<m$.
If $K$ is a Kakeya set in $F_q^m$ then $K\times F_q^{n-m}$ is a Kakeya set in $F_q^n$.

Attempt
Actually this is true.
Suppose that $K\times F_q^{n-m}$  isn´t a Kakeya set in $F_q^n$ then we must have
$$|K\times F_q^{n-m}|<\frac{q^n}{n!}$$
But notice that
$$|K\times F_q^{n-m}|=|K||F_q^{n-m}|=|K|q^{n-m}$$
Since $K$ is a Kakeya set in $F_q^m$ then $|K|\geq \frac{q^m}{m!}$ and $m!<n!\rightarrow \frac{1}{n!}<\frac{1}{m!}$ Therefore
$$|K||F_q^{n-m}|=|K|q^{n-m}\geq \frac{q^m}{m!}q^{n-m}\geq \frac{q^n}{n!}$$ a contradiction.
Then $K$ is a Kakeya set.
Is the proof right? if it had a mistake where and how can i prove that.
Note:
Let $q$ the power of a prime number $p$, consider the finite field $F_q^n$, A set $K\subset F_q^n$ is a Kakeya set if for every direction $v\in F_q^n\setminus \lbrace 0 \rbrace$ the line $L_{v,a}\subset K$
 A: I haven't checked any sources, but presumably the correct definition of a Kakeya set is the following.

$K\subseteq \Bbb{F}_q^m$ is a Kakeya set if to each vector $v$ in $\Bbb{F}_q^m\setminus\{0\}$ (= a direction) there exists a vector $a\in\Bbb{F}_q^m$ such that the line
$$L_{v,a}=\{a+tv\mid t\in\Bbb{F}_q\}\subseteq K.$$

Apparently $|K|\ge q^m/m!$ is a known lower bound to the size of a Kakeya set $K$. A fundamental mistake in your attempt is that you treat this lower bound (a necessary condition) as a sufficient condition. After such a logical mistake no recovery is possible.
Assuming I guessed the correct defition of a Kakeya set (please provide one!), the claim is straight forward using
The observation: A vector $(x,y)\in \Bbb{F}_q^m\times \Bbb{F}_q^{n-m}$ is an element of $K\times \Bbb{F}_q^{n-m}$ if and only if $x\in K$.
The main claim follows from this immediately. Let us fix a direction $v=(v_1,v_2)\neq0$, where $v_1\in\Bbb{F}_q^m, v_2\in\Bbb{F}_q^{n-m}$.

*

*If $v_1\neq0$ then by the definition of a Kakeya set there exists a vector $a_1\in\Bbb{F}_q^m$ such that $a_1+tv_1\in K$ for all $t\in\Bbb{F}_q$. Let us choose $a=(a_1,0)\in \Bbb{F}_q^m\times\Bbb{F}_q^{n-m}$. By the observation all the vectors $a+tv=(a_1+tv_1,tv_2)$ are elements of $K\times\Bbb{F}_q^{n-m}$. Hence this line $L_{a,v}$ works.

*If $v_1=0$ then we can select any point $a_1\in K$, and the line $L_{a,v}$ with $a=(a_1,0)$ works.

Hence $K\times \Bbb{F}_q^{n-m}$ is a Kakeya set.
