Every few years this question re-appears in my life (as it just did this week) and I have to rediscover the Euler Characteristic proof which I half-remember. Let me just write my thoughts down here so I don't ever have to do this again.
Definition. Let $A$ be a non-zero commutative ring. A finite free resolution (or FFR) of an $A$-module $M$ is an exact sequence $$0\to F_n \to F_{n-1} \to \cdots \to F_1 \to F_0 \to M\to 0$$ where each $F_i$ is a free $A$-module, of rank $r_i$.
If $M$ has an FFR then we define the Euler Characteristic $\chi(M)$ of $M$ to be $\sum_i(-1)^ir_i$. This turns out to be a well-defined invariant of $M$ (i.e. independent of the resolution) because a relatively straightforward induction on $n$ shows that if $G_i$ is another resolution then $F_0\oplus G_1\oplus F_2\oplus\cdots\oplus (F\ \mbox{or}\ G)_n$ is isomorphic to $G_0\oplus F_1\oplus G_2\oplus\cdots$. For more details on this part of the argument see Lemma 4 in section 19 of Matsumura's book "Commutative Ring Theory".
The fundamental fact is:
Theorem. If $M$ has an FFR then $\chi(M)\geq0$.
As a consequence, not only does an injection $A^m \to A^n$ imply $m\leq n$ (apply the theorem to the cokernel), but if $0\to A^m\to A^n \to A^r$ is exact then $m+r\geq n$ and so on and so on. Do the other solutions also generalise to prove this?
The theorem is Theorem 19.7 of Matsumura's "commutative ring theory". Here's a sketch. Localising at a minimal prime one can assume that $A$ is local with nilpotent maximal ideal. It suffices to prove that in this case $M$ is free, as then the exact sequence splits and the Euler Characteristic is just the rank of $M$. It suffices to prove that the cokernel of $F_n\to F_{n-1}$ is free, because then one is done by induction on $n$. The trick now is to replace $F_n$ and $F_{n-1}$ by a minimal free resolution, look at the image of $F_n$ and see that its coordinates must lie in the maximal ideal, and it's then not difficult to concoct a non-zero element of $A$ which kills this finite set of coordinates, and hence $F_n$ is a free module annihilated by a non-zero element of $A$ and must hence be zero.
This last part of the argument feels very similar to Georges Elencwajg's argument above, which reduces to the Noetherian case and uses lengths, but Matsumura explicitly avoids this reduction.
