A question on angular momentum for a body with constant velocity

Let a particle be moving in space with constant velocity. We are required to show that for that particle , angular momentum is constant throughout the motion irrespective of origin we choose.

MY PROOF:-

Let's choose an origin $$O$$ and $$3$$-dimensional Cartesian coordinate system.

Let the position vector of required particle from origin be $$\vec{r}$$ and constant velocity vector at that point be $$\vec{v}$$. And let it's mass be $$m$$ then angular momentum is $$\vec{L}=\vec{r}×m\vec{v}$$

Differentiating on both sides, $$\frac{d\vec{L}}{dt}=\frac{d\vec{r}}{dt}×m\vec{v}+m\vec{r}\frac{d\vec{v}}{dt}$$

Since the velocity is constant $$\frac{d\vec{v}}{dt}$$ term goes to $$0$$. The first term obviously goes to $$0$$ as $$\vec{v}=\frac{d\vec{r}}{dt}$$. So, $$\frac{d\vec{L}}{dt}=0$$ So $$\vec{L}$$ is constant. Since the origin we chose was arbitrary, it works at any position.

Hence proved.

I was just wondering if my solution is true.

You can also show that $$\vec{r} = \vec{r}_0 + \vec{v}\, t$$ and that
$$\require{cancel} \vec{L} = \vec{r} \times m \vec{v} = \vec{r}_0 \times m \vec{v} + ( \cancel{ \vec{v} \times m \vec{v} }) t = \text{(const)}$$