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, $$\boxed{v = ω \times r}$$
I'll try to give here a very partial answer to the question. Not sure whether it is interesting in itself, but it might provide a hint for further development. Possibly its place would be within a comment, but comments are limited in length and it wouldn't fit.
Let's define $\alpha (t_0,t)$ as the $\alpha (t)$ from the question related to the given $t_0$. We identify the rotation vector with the rotation itself. For $t_i = t_0 + i\ dt$ and $t = t_0 + n\ dt$, we have by simple composition of successive rotations
$$\alpha (t_0,t) = \prod_{i=n-1}^{0} \alpha (t_i,t_{i+1}).$$
We know from this question/answer that $\partial_2 \alpha (t,t) = \omega(t)$ for all $t$. Using $\alpha (t,t+dt) = I+\partial_2 \alpha (t,t)\ dt + o(dt) = I+\omega(t)\ dt + o(dt)$, we have
$$\alpha (t_0,t) = \prod_{i=n-1}^{0} I+\omega(t_i)\ dt + o(dt).$$
This provides by the way a numerical method expressing $\alpha$ in terms of $\omega$.
In the very special case where all these rotations commute (example of common axis), the limit is an exponential: taking the log and the limit when $dt \to 0$, we have
$$\log (\alpha (t_0,t)) = \sum_{i=n-1}^{0} \omega(t_i)\ dt + o(dt) \to \int_{t_0}^{t} \omega(s) \ ds,$$
hence
$$\alpha (t_0,t) = \exp \left( \int_{t_0}^{t} \omega(s) \ ds \right).$$
In the general non commutative case, the log will involve Lie brackets starting with $dt^2[\omega(t_i),\ \omega(t_j)]$ (cf. Dynkin's formula), and some more courage seems required.
EDIT: according to a comment below from Keshav Srinivasan, the expression above becomes in the general non commutative case
$$\alpha (t_0,t) = \operatorname{OE}[\omega](t_0,t)= \mathcal{T} \left\{e^{\int_{t_0}^{t} \omega(s) \, ds}\right\},$$
see Ordered exponential for the definition of $\operatorname{OE}[\omega](t)$. Though this is not an exact answer to the question, as it involves rotation matrices instead of the required rotation vectors.
Best Answer
Set your copy of Halliday, Walker & Resnick on the table so the front cover is parallel to the table and visible to you. Now your right hand flat on the book with your thumb and forefinger forming a ninety degree angle. Orient your hand so your thumb is parallel to the spine of the book and pointing toward the top edge. Your forefinger should be parallel to the lines of text on the cover, pointing to the right.
This makes a nice basis for a book-based coordinate system. Your thumb points along the x-hat axis, your forefinger along the y-hat axis. To complete a right-hand system, the z-hat axis points into the book.
Pick the book up and make a +90 degree rotation about the x-hat axis. The spine of the book should be horizontal and facing up. Now make a +90 degree rotation about the y-hat axis (as rotated by that first rotation). You should be looking at the front cover of the book but oriented vertically with text flowing toward the ground.
If you start all over again but reverse the order of operations, +90 degree rotation about the y-hat axis followed by a +90 degree rotation about the x-hat axis, you should be looking at the spine of the book rather than the front cover. The front cover is oriented vertically, but with text running parallel to the ground. You can put your book back in the bookcase.
Rotation in three dimensional space and higher is not commutative (rotation A followed by rotation B is not necessarily the same as rotation B followed by rotation A).
Another hint that there's something different between rotation and translation is the number of parameters needed to describe the two in some N-dimensional space. Translation in N dimensional space obviously needs N parameters. Lines have one degree of freedom, planes, two, three dimensional space three, and so on. Lines don't rotate. There are no rotational degrees of freedom in one dimensional space. Rotation does make sense in two dimensional space, where a single scalar (one rotational degree of freedom) completely describes rotations. Three dimensional space has three degrees of freedom. Four dimensional space? It has six. Three dimensional space is the only space for which the number of rotational degrees of freedom and number of translational degrees of freedom are equal to one another.
This unique characteristic of three dimensional space is why you can treat angular velocity as a vector. The vector cross product (something else that is unique to three dimensional space) means introductory students can be taught about rotations without having to learn about Lie theory or abstract algebras. That wouldn't be the case for students who live in a universe with four spatial dimensions.