We define a polynomial trend model as \begin{align} X_t = a_n t^n + \ldots + a_1 t + a_0 + Z_t = \sum_{j=0}^n a_j t^j + Z_t, \qquad (*) \end{align} for a strictly stationary time series $Z_t$. We define the seasonal difference filter as \begin{align} \nabla_k\ X_t = X_t - X_{t-k}, \end{align} where $k$ is the length of the season or period. Now, I am wondering if a (repeated) seasonal filter will remove a polynomial trend, i.e. $X_t$ becomes a stationary process.
In order to make this question more specific, it could be shown that \begin{align} \nabla_k^n\ X_t = \sum_{i=0}^n {n \choose i}(-1)^i\ X_{t-ik}. \end{align} And implementing $(*)$ in the above expression yields \begin{align} \nabla_k^n\ X_t &= \sum_{i=0}^n {n \choose i}(-1)^i \sum_{j=0}^n a_j (t-ik)^j + Z_{t-ik} \\ &= \sum_{i=0}^n {n \choose i}(-1)^i \sum_{j=0}^n a_j \sum_{l=0}^j {j \choose l} t^j (-ik)^{j-l} + Z_{t-ik}. \end{align} For $n=1$ the statement holds, for $k=1$ $X_t$ becomes stationary.
To continue the proof by induction, for $n=m$ we assume that $X_t$ could become stationary for some $k \in \mathbb{N}$.
However, for $n = m+1$, how to find back the expression of $n=m$ in \begin{align} \sum_{i=0}^{m+1} {m+1 \choose i}(-1)^i \sum_{j=0}^{m+1} a_j \sum_{l=0}^j {j \choose l} t^j (-ik)^{j-l} + Z_{t-ik}\ ? \end{align} I am aware of the fact that the sums could be elaborated in terms of $1\leq i \leq m$ and $m+1$ and that \begin{align} {m+1 \choose i} = \frac{m+1}{m+1-i} {m \choose i}. \end{align} However, I do not see how to conclude the proof. Is there any other approach to prove this statement?