I am trying to prove that linear difference operator, $(σ-1)^{k+1} (p) = 0$ for all $p$ $\epsilon$ $\mathbb{Q}[t]$, with $deg(p) \leq k$.
In this case $\sigma(t)=t+1$ and $\sigma($anything else$)=$ whatever it was.
I tried induction, but I wasn't sure how to handle $p$. So for base case I did $k = 0$, therefore $p$ has to be a constant, $c$, since $deg(p) \leq 0$.
Then $(\sigma -1)^{0+1}(c)=c-c=0$
But I was confused on how to do the induction step exactly since I need to account for all $p$, such that $deg(p) \leq k+1$.
I could say $(\sigma-1)^{k+2}(p) = (\sigma-1)(\sigma-1)^{k+1}(p)$ and go from there, but I'm not sure if I can do it with the same $p$ or not. Any help would be appreciated! Thanks!
Hint: Show that $(\sigma-1)p$ has lower degree than $p$. And to show this, show that $(\sigma-1)t^m$ has degree $<m$ and then use that $\sigma$ is linear.