I am reading "Computing the Continuous Discretely" by Mattias Beck and Sinai Robins and am working on the following exercise (3.13) which is to prove the following:
If $$\sum_{t\geq 0}f(t)z^t = \frac{g(z)}{(1-z)^{d+1}},$$
then $f$ is a polynomial of degree d if and only if $g$ is a polynomial of degree at most $d$ and $g(1)\neq 0$.
I think that if $deg(g) > d+1$, then the LHS is an improper rational function and could be written as a quotient and remainder, that is something that is not of the form of the RHS. I also know that $$\frac{1}{(1-z)^{d+1}} = \sum_{n=0}^{\infty} {n + d \choose d}z^n.$$ Since $f(t)$ is not a polynomial in $z$, I am guessing that the condition requiring $deg(f) = d$ might be related to the above identity although I'm not really sure how to proceed from here.
As you correctly mentioned, there is an expansion $$ q(z) = \frac{1}{(1-z)^{d+1}} = \sum\limits_t \binom{t+d}{d} z^t. $$ If $g(z) = g_0 + g_1 z + \dots + g_d z^d$, it means that, generally,
$$\begin{align} f(t) &= [z^t] g(z) q(z) \\ &= [z^t] \sum\limits_{i=0}^d g_i z^i q(z) \\ &= \sum\limits_{i=0}^d g_i [z^{t-i}] q(z) = \boxed{\sum\limits_{i=0}^d g_i \binom{t-i+d}{d}} \end{align} $$ The later expression is a polynomial in $t$ of degree at most $d$. Note that any polynomial $f(t)$ of degree at most $d$, in turn, can be represented as such linear combination, that is, it is always possible to find $g_0, \dots, g_d$ for which the identity holds (see falling factorial basis and Stirling numbers of the second kind).
The property $g(1) \neq 0$ is same as saying that $g(z)$ is not divisible by $z-1$ due to little Bézout's theorem. In other words, it means that the fraction can not bet reduced and $d$ is the minimum possible for which it can be represented like this. This implies that the bound on $d$ is tight, thus $d$ is exactly the degree of $f(t)$.
P.S. There are, of course, other, more insightful, ways to show it. For example, this generating function identity highlights the fact that any polynomial of degree $d$ is a linear recurrence with characteristic polynomial being simply $(x-1)^{d+1}$. Yet another possible way to prove the result in the question is by taking finite differences of $f(t)$ for $d$ times and analyzing the result.