Let $G$ be a finite a $p$-solvable group, where $p$ is a prime number, (that is there is a normal series in $G$: $1=N_0<N_1<\ldots<N_r=G$ such that $N_i/N_{i-1}$ is a group of order not divisible by $p$ ($p'$-group) or $p$-group). The minimal number of factors in any such series, which are $p$-groups, is called $p$-length of $G$.
I try to show that $p$-lenth of $G$ is less or equal the nilpotence class of the Sylow subgroup $P$ of $G$. I will be grateful for the ideas and guidance.
To further the hint: it follows from the fact mentioned in comments that for a general (finite) $p$-solvable group $G$ with Sylow $p$-subgroup $P \neq 1$, we have $Z(P) \subseteq O_{p^{\prime},p}(G)$ so that a Sylow $p$-subgroup of $G/O_{p^{\prime},p}(G)$ has nilpotency class strictly smaller than that of $P$.