OK, I'm assuming you want the formal proof of this well known kinematics formula! So here goes:

Let the particle rotate about the axis OO' ... Within time interval dt let its motion be represented by the vector dφ whose direction is along axis obeying right-hand-corkscrew rule, and whose magnitude is equal to the angle dφ.

Now, if elementary displacement of particle at a be specified by radius vector r,

From the diagram, it is easy to see that, for infinitesimal rotation, dr= dφΧr ... 1 (crossproduct)

By definition, ω = dφ/dt

Thus taking the elementary time interval as dt, all given equations surely hold!

Thus we can divide both sides of equation 1 by dt which is corresponding time interval!

So we get dr/dt = dφ/dt X r of course r value won't change WRT the particle and axis, so r/dt is essentially r!

So result is, v = ωΧr

OK, I'm assuming you want the formal proof of this well known kinematics formula! So here goes:

Let the particle rotate about the axis OO' ... Within time interval $dt$ let its motion be represented by the vector $d\varphi$ whose direction is along axis obeying the right-hand-corkscrew rule, and whose magnitude is equal to the angle dφ.

Now, if the elementary displacement of particle at a be specified by radius vector $r$,

From the diagram, it is easy to see that, for infinitesimal rotation, $dr= d\varphi\times r \tag{1}$

By definition, $ω = dφ/dt$

Thus taking the elementary time interval as $dt$, all given equations surely hold!

Thus we can divide both sides of the equation $(1)$ by $dt$ which is the corresponding time interval!

So we get $dr/dt = dφ/dt \times r$ of course $r$ value won't change WRT the particle and axis, so $r$/dt is essentially $r$!

So result is, $$\boxed{v = ω \times r}$$