prove this $x^{m+1}|f^{(m)}(x)-x$ Let $f(x)$ is polynomial with complex coefficients,such
$$x^2|f(x)-e^{\frac{2\pi i}{m}}\cdot x$$
where $m>1$ be give postive integers,and define 
$$f^{(1)}(x)=f(x),f^{(2)}(x)=f(f(x)),f^{(3)}(x)=f(f(f(x))),\cdots,f^{(m)}(x)=f(f^{(m-1)}(x))$$
show that
$$x^{m+1}|f^{(m)}(x)-x$$
I try to use mathematical induction to prove.But there is no proof
 A: Consider the space $A = t\Bbb C[[t]] = \{a_1t + a_2t^2 + a_3t^3 + \ldots \mid a_i \in \Bbb C\}$.
$A$ can be equipped with the composition law $\circ$ which is linear in the first argument
( $f \circ h + g \circ h = (f+g) \circ h$ and $(\lambda f) \circ g = \lambda (f \circ g)$ for $\lambda \in \Bbb C$). 
Thus for any $g \in A$, we get a linear endomorphism $\rho_g(f) = f \circ g$.  
If $g = b_1t + b_2t^2 + b_3t^3 + \ldots$, then $\rho_g(t^k) = g(t)^k = b_1^k t^k + \ldots$.
This shows that the "matrix" of $\rho_g$ is triangular (in particular, $\rho_g$ is compatible with the $t$-adic topology) and the coefficients on the diagonal are the sequence $(b_1^n)$.
Restricting modulo $t^{n+1}$ gives you a linear map $\rho_g^{[n]} : A_n \to A_n$ (where $A_n = t\Bbb C_{n-1}[t]$ has dimension $n$) defined by $\rho_g^{[n]}(f) = \rho_g(f) \pmod {t^{n+1}}$ whose matrix is simply the $n \times n$ submatrix in the topleft corner of the infinite matrix of $\rho_g$. Its eigenvalues are the $b_1^k$ for $k=1 \ldots n$.
Now suppose you are looking at a $g$ whose $b_1$ is a primitive $n$th root of unity $\zeta_n$.
Then $\rho_g^{[n]}$ has eigenvalues $\zeta_n^k$ for $k=1 \ldots n$, and since they are all distinct this is diagonalisable, and since their $n$th power is $1$, $(\rho_g^{[n]})^n$ is the identity of $A_n$.
Going back to $A$, this proves that the topleft $n \times n$ block in the matrix of $\rho_g^n$ is $I_n$, and so that for any $f \in A$, $f \circ g^{\circ n} \equiv f \pmod {t^{n+1}}$
Applying this to $t \in A$ you get $g^{\circ n} \equiv t \pmod {t^{n+1}}$
