Why is the norm of an ideal contained in that ideal?

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Suppose $K$ is a number field and that $\mathcal{O}_K$ is the ring of integers of $K$. Now, let $I$ be an ideal in $\mathcal{O}_K$. I know that $N(I) \in I$, but I want to prove it.

By definition, $N(I)= \#\mathcal{O}_K / K$.

A proof given by our lecturer invokes "a theorem of Lagrange" to note that, for any $x \in \mathcal{O}_K$ we have that $N(I).x \in \mathcal{O}_K$ and so by setting $x=1$ the result follows.

Can someone please flesh this out a little for me.

What is the theorem of Langrange (because I can't see how 'Langrange's Theorem' gives us this result)? And how is it applied?

Thanks.