We want to prove the Chinese Remainder Theorem, and for s>2 we got to the point that:
$$R/\left(\left(I_1\cap ...\cap I_{s-1}\right)\cap I_s\right)=R/\left(\left(I_1\cdot ...\cdot I_{s-1}\right)+I_s\right)\:\simeq \:R/\left(I_1\cdot ...\cdot I_{s-1}\right)\:\times \:R/I_s\:=\:\left(R/I_1\:\times \:...\:\times \:R/I_{s-1}\right)\times R/I_s$$
I understand that the last equal sign comes from our induction assumption (as we proved it for s=2 already).
What I don't get is that what we wanted to also show (besides the isomorphism) is that $$I_1\cdot ...\cdot I_s=I_1\cap ...\cap I_s$$ but we get
$$\left(I_1\cdot ...\cdot I_{s-1}\right)+I_s=I_1\cap ...\cap I_s$$
apparently
So where's the mistake here? We already showed that
$$\left(I_1\cdot ...\cdot I_{s-1}\right)+I_s = R$$
Because $$1\in \left(I_1+I_s\right)...\left(I_{s-1}+I_s\right)\subset \left(I_1\cdot ...\cdot I_{s-1}\right)+I_s$$