suppose $G$ is a finite solvable group,then $G$ is nilpotent if and only if all Hall subgroups of $G$ which its indices are power of a prime number are normal.
suppose $G$ is a solvable finite group and $|G|=p_1^{\alpha_1}p_2^{\alpha_2}...p_n^{\alpha_n}$ that $p_1,p_2,...,p_n$ are prime numbers.
now suppose $G$ is nilpotnt,if all sylow subgroups of it is normal if $P_1 \in Syl_{p_1}(G),P_2 \in Syl_{p_2}(G),...,P_{n-1} \in Syl_{p_{n-1}}(G)$,then $P_1,P_2,...,P_{n-1}\triangleleft G$ but $P_1,P_2,...,P_{n-1}$ is one for subgroup of $G$ from order $p_1^{\alpha_1}p_2^{\alpha_2}...p_{n-1}^{\alpha_{n-1}}$ that its indices in $G$ is $p_n^{\alpha_n}$ ,and what we want will achieved with continuing this process.
for the inverse I should show that $G$ is nilpotent,I think I must show that all sylow subgroups are normal but I don't know what to do and I stuck,please help me to make the inverse complete.thanks.
I think this follows from a Theorem of P. Hall saying that in a solvable group $G$ of order $p_1^{a_1}p_2^{a_2}\cdots p_n^{a_n}$ there do exist subgroups $H_i$ of order $\vert G\vert/{p_i^{a_i}}$ (these are ``Hall-$\pi$-subgroups'' for $\pi=\{p_1,\dots,p_{i-1},p_{i+1},\dots,p_n\}$). According to your assumption each $H_i$ is normal. So one can write $G=H_iP_i$ and $G/H_i\cong P_i$. Doing this for each $i$ one gets a morphism $G\to P_1\times P_2\times \cdots\times P_n$ whose kernel is $H_1\cap H_2\cap \cdots\cap H_n=1$, so it must be an isomorphism.
The other direction is incorrect and the statement should be reformulated. Take, for example, some prime $p$ and a $p$-group $G$ which does have a non-normal subgroup. Then the index of this subgroup is a prime-power but the subgroup is not normal.