In every curve $\alpha$ parametrized by arc length, $\alpha''(s) \cdot \alpha'(s) = 0$ In my book, in the first step of the proof they say the following, where $\cdot$ means dot product:

Since $\alpha$ is parametrized by arc length, we have
  $$1 = |\alpha'(t)|^2 = (\alpha'(s) \cdot\alpha'(s))$$
  for all t

This is not at all clear to me. I don't see why that statement is true and have no idea how to prove it, and am confused that there are two different variables...
 A: I'll address the question expressed in the comment. If
$$ s(t) = \int_{a}^t | \alpha'(u) | \, du $$
is the arclength of the curve $\alpha$ (from $\alpha(a)$ to $\alpha(t)$) then by the fundamental theorem of calculus and the chain rule, we have
$$ \frac{ds}{dt} = |\alpha'(t)|, \frac{dt}{ds} = \frac{1}{\frac{ds}{dt}} =  \frac{1}{|\alpha'(t(s))|}. $$
If we consider the curve $\alpha(t(s))$ which is parametrized by arclength and take the derivative using the chain rule, we get
$$ \frac{d}{ds}(\alpha(t(s))) = \alpha'(t(s)) \frac{dt}{ds} = \frac{\alpha'(t(s))}{|\alpha'(t(s))|} $$
which has unit length so the derivative of $\alpha$ with respect to $s$ has unit length.
A: I think the $t$ in your statement should be $s$. As written it's confusing.
In any parameterization the vector $\alpha '(t)$ is the tangent vector to the curve at point $\alpha(t)$. In a short time interval the position changes by $\alpha '(t) dt$; you integrate $|\alpha '(t)| dt$ with respect to $t$ to compute the arclength up to $t$.
If $t$ is the parameterization by arclength then the arclength up to $t$ is just $t$ so the integrand $|\alpha '(t)|  $ must be identically $1$.
The last step is just the general fact that for any vector $v$
$$
|v|^2 = v \cdot v .
$$
