If $a_{g}(1)=g(x)$ and $a_{g}(r)=g(a_{g}(r-1))$ for $r>1$ then is it true that $\limsup\limits_{r\to\infty}\gamma(a_{r})=\infty?$ Let $g(x)$ be a polynomial with integral coefficients.We define $\gamma(g(x))$ to be the degree of the non constant  polynomial $r(x)$ which  divides $g(x)$ for all $x$ and also has minimal degree.
For $r\geq 1$, We define the sequence $a_{g}$ for some polynomial $g(x)$ as follows:
$\clubsuit)a_{g}(1)=g(x)$
$\clubsuit)a_{g}(r)=g(a_{g}(r-1))$ for $r\geq 2$
Now we are given a polynomial $f(x)$ of $\deg(f)\geq 2$ such that if $x=0$, $a_{f}(r)\rightarrow\infty$ as $r$ grows large.
Then is it true that
$$\limsup_{r\to\infty}\gamma(a_{f}(r))=\infty?$$
 A: Yes, it is true, and I'm guessing it's well-known. (For example, it might be a theorem in Chapter 3 of Silverman's "The Arithmetic of Dynamical Systems", but I don't own that book yet.)
It is useful to use the terminology of arithmetic dynamics. We call the polynomials $a_{f}(r)$ the iterates of $f$, and for a complex number $z$, we call $a_{f}(r)(z)$ the $r$th iterate of $z$. We call the set $\{ 0, f(0), f(f(0)), \ldots \}$ the forward orbit of $0$ (which by assumption tends to infinity). An $r$th preimage of $0$ is a number $\beta_{r}$ with $a_{f}(r)(\beta_{r}) = 0$. The assumption that the forward orbit is infinite implies that there are infinitely many preimages of $0$. For if $\beta_{r_{1}}$ is an $r_{1}$th preimage and $\beta_{r_{2}}$ is an $r_{2}$th preimage, then $\beta_{r_{1}} = \beta_{r_{2}}$ implies that $r_{1} = r_{2}$ as can be seen by considering the $\max(r_{1},r_{2})$ iterates of $\beta_{r_{1}}$ and $\beta_{r_{2}}$. One of these is zero, and one of them is in the forward orbit of $0$. If these are equal, then $0$ is a periodic point, contradicting that the forward orbit of $0$ is infinite.
Suppose to the contrary that $d = \limsup_{r \to \infty} \gamma(a_{f}(r)) < \infty$. Then, there are infinitely many preimages of $0$ that lie in extensions of $\mathbb{Q}$ of degree $d$. For a point $P$ in $\mathbb{P}^{n}(\overline{\mathbb{Q}})$, let $H(P)$ denote the absolute (non-logarithmic) height of $P$. Theorem 5.6 of Silverman's Arithmetic of Elliptic Curves states that if $F : \mathbb{P}^{N} \to \mathbb{P}^{M}$ is a morphism of degree $d$, there are positive constants $C_{1}$ and $C_{2}$ (depending on $F$) so that
$$ C_{1} H(P)^{d} \leq H(F(P)) \leq C_{2} H(P)^{d}. $$
To apply this, we let $F : \mathbb{P}^{1} \to \mathbb{P}^{1}$ be the morphism induced from the polynomial $f$ (given by $F((x : 1)) = (f(x) : 1)$ and $F((1 : 0)) = (1 : 0)$). It is straightforward to see that for any $r$th preimage $\beta_{r}$ of $0$, we have that the height of $\beta_{r}$
$$ H((\beta_{r} : 1)) \leq \max(|f(0)|,1) \cdot \left(\frac{1}{C_{1}}\right)^{\frac{1}{\deg(f) - 1}} $$
is bounded.
Theorem 5.11 of Silverman's Arithmetic of Elliptic Curves shows that the set of algebraic points in $\mathbb{P}^{N}$ with height $\leq \max(|f(0)|,1) \cdot \left(\frac{1}{C_{1}}\right)^{\frac{1}{\deg(f) - 1}}$ and lying in a number field of degree $\leq d$ is finite. It follows that there are only finitely many preimages of $0$. This contradicts the first paragraph.
(With more work, one can show that the assumption that $d = \limsup_{r \to \infty} \gamma(a_{f}(r))$ implies that there is a fixed number field $K/\mathbb{Q}$ of degree $d$ that contains infinitely many preimages of $0$.)
