If f is in Schwartz Space, then derivative is in $L^p$

Question

In Folland's Real Analysis chapter 8.1, there is a remark about Schwarz Space $S$: ''It is an important observation that if $f\in S$ , then $∂^αf\in L^p$ for all α and all $p \in [1,∞].$ Indeed, $|∂^αf(x)|≤C_N(1+|x|)^{−N}$ for all N , and $(1+|x|)^{−N}∈L^p$ for $N>n/p$ by Corollary 2.52'' The corollary 2.52 states that if $C,c$ are constants and B={$x\in R^n:|x|<c$} the following statements hold (a.)$|f| \leq C|x|^{-a}$ for $a<n$, then $f \in L^1(B)$ also (b.)$|f| \leq C|x|^{-a}$ for $n<a$, then $f \in L^1(B^c).$ I'm trying to use the corollary rigorously to see that result, but it isn't obvious to me how to apply the corollary as it separates the cases $a<n$ and $a>n$. I don't understand why we are just taking the case $N>n/p$ to conclude that $∂^αf(x) \in L^p$. Can someone clarify it?