Characterization of nilpotency with normal subgroups

I have to prove the following:

Let G be a finite group then G is nilpotent $\iff$ every proper normal subgroup, $N\lhd G$, satisfy $[N,G]\leq N$

But it's not true that for every normal subgroup (even if G is not nilpotent) $[N,G]\subseteq N$ as $ngn^{-1}g^{-1}\in N$?

I found the solution in this book. And as Maths Rahul pointed $N$ has to be a proper subgroup.

We prove it by induction over the order of $[N,G]$, if $[N,G]=1$ trivialy $[N,G]<N$.

If $[N,G]>1$ as $G$ is nilpotent we have $N\cap Z(G)\neq1$ for every normal subgroup $N$ of $G$ and $[N,G]\lhd G$ we have $[N,G]\cap Z(G)=K\neq1$. Taking the quotient $$\frac{N}{K}\lhd\frac{G}{K}$$ by induction $$\frac{[N,G]}{K}<\frac{N}{K}$$ and using the correspondence $[N,G]<N$

The statement is not true. Just consider any non-Abelian simple finite group, such as $A_5$. It does not have proper normal subgroups, so each of them satisfies your condition.

The fact that you mentioned: "For every normal subgroup $N$ $[N,G]\subseteq N$ as $ngn^{-1}g^{-1}\in N$" $(n\in N, g\in G)$" is true because $ngn^{-1}g^{-1}=n(gn^{-1}g^{-1})$ and $gn^{-1}g^{-1}$ is in $N$ since $N$ is normal.