Noetherian module does not contain a submodule $N$ which is a direct sum of $n$ simple modules

The question:

Let $$R$$ be a ring and $$M$$ be a Noetherian module. Prove there is $$k \in \mathbb{N}$$ such for all $$n > k$$, $$M$$ does not contain a submodule $$N$$ which is a direct sum of $$n$$ simple modules.

My thoughts: Since M is Noetherian, this means any submodule and factor module of M is Noetherian. I considered a submodule, $$N$$. Since N is noetherian, N satisfies the ascending chain condition. Is this where I pick my k? I am not sure about other properties on Noetherian modules.

(i) There's a finite number of isomorphism classes of simple modules that appear as submodules of $$M$$ (because $$M$$ is noetherian)
(ii) Given a simple module $$N$$, the natural numbers $$n$$ such that $$N^n$$ is (isomorphic to) a submodule of $$M$$ are bounded : this is the "hard" bit : consider an increasing sequence $$n_i$$ of such $$n$$'s, let $$N^{n_i}\simeq L_i \subset M$$ and consider $$\displaystyle\sum_i L_i$$, which is finitely generated, therefore there is a finite amount of $$L_i$$'s such that any $$L_j$$ is in their sum. Take $$n_j > \sum_i n_i$$ where the sum runs over this finite amount of $$i$$'s
Now by induction prove that any finite sum of submodules of $$M$$ isomorphic to $$N$$ is isomorphic to a single $$N^p$$ for some $$p$$ smaller than the number of submodules; thus for this $$j$$ we have an injection $$N^{n_j}\to N^p$$ for some $$p; and that's not possible (indeed it is represented by an $$n_j\times p$$ matrix over the division ring $$\hom_R(N,N)$$ and such a matrix necessarily has a nonzero vector $$(f_1,...,f_{n_j})$$ in its kernel by dimension theory, thus if you apply it to some $$n\in N$$ such that $$(f_1(n),...,f_{n_j}(n))\neq 0$$ you get a nonzero element of the kernel of $$N^{n_j}\to N^p$$)
(iii) You can then find $$k$$ to be the sum over the finite number of isomorphism classes of simple modules of the bound given by (ii)