Polynomials $f$ such that $f(2t)=h(f(t))$ for some $h$ From Barbeau's Polynomials

  
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*Find all polynomials $f$ such that $f(2t)$ can be written as a polynomial in $f(t)$, i.e. for which there exists a polynomial $h$ such that  
$f(2t)=h(f(t))$

When I went for the answer, I found:   

When I went to the answer, I couldn't understand it, can you help me? I'm trying to know what's he doing in this answer, I guess it's a way to prove it, but It's still intractible to me. You can explain me or recommend me some thing for reading.
Thankyou.
 A: If you have $f(t)$ and $h(t)$ of degree $a$ and $b$ respectively, then the degree of $f(h(t))$ and $h(f(t))$ should be ab. Because the highest degree term should be $(x^a)^b=(x^b)^a=x^(ab)$.
So by that, $deg(f(t))=deg(f(2t))=deg(f)*deg(h)$
$deg(f)*(deg(h)-1)=0$
So $deg(f)=0$ ($f(t)$ is a constant function) or $deg(h)=1$ ($h(t)$ is linear) or both.
Then the proof just let $h(t)=ut+v$ and try equating $f(2t)=h(f(t))$ and comparing like-coefficients. 
Note that it is impossible for $u=2^k$ for all positive integers $1\leq k\leq n$, because $u$ can only has one single value. But $f(t)$ is assumed to have degree $n$, so $u$ must be equal to $2^n$, otherwise the term $a_n$ must equal $0$, contradicting the fact that $deg(f)=n$.
Then you have a general form for $f(t)$ and $h(t)$, by the assumption that both of the functions are polynomials (thus have finite degree).
For part b, observe that $deg(h(t))=2$, and $f(t)$ is not constant. That contradicts our result that $deg(h)=1$, so $f(t)$ must not be a polynomial.
