If $I$ is finitely generated nilpotent and $R/I^{n-1}$ is noetherian then $R$ is noetherian 
If $I$ is a finitely generated ideal of a commutative ring $R$ with $1$ such that $I^n = \{0\}$ and $R/I^{n-1}$ is noetherian, then $R$ is also noetherian.

I don't know what I should do. If I can prove for example that $I^{n-1}$ is finitely generated as $R/I^{n-1}$ module then I can deduce that it is noetherian as $R$-module and then I am done. But I don't know if that is true and how to prove it.
Thanks.
Added: also isn't it possible to prove that $I^{n-1}$ is finitely generated without requiring $I^n = \{0\}$? (just realized that)
I mean I can prove that if $I$ and $J$ are finitely generated ideals then $IJ$ is also finitely generated so by induction if $I_1,...,I_k$ are finitely generated then $I_1...I_k$ is so. So $I^{n-1}$ is finitely generated as $R/I^{n-1}$ module so it is noetherian as $R$-module so $R$ is then noetherian. Is there something wrong with this? 

Since $I^n=0$ the ideal $I^{n-1}$ is a finitely generated $R/I^{n-1}$-module: it is finitely generated (any power of a finitely generated ideal is finitely generated) and $I^{n-1}\cdot I^{n-1}=0$; see also here. Now use the exact sequence of $R$-modules $$0\to I^{n-1}\to R\to R/I^{n-1}\to 0.$$
