Linderholm's book, Mathematics Made Difficult, is full of this sort of thing. These samples might give you the idea:
- Proposition. If you can add, you can count.
Proof. In counting with the additive monoid $N_0$, we start at $0$; after saying $n$, we say $n+1$. Thus, we already have a counting system. $$n\mapsto n+1\qquad\{0\}\hookrightarrow N_0\rightarrow N_0.$$ But having a counting system is not enough. What you must have in order to assure success in all your counting endeavours is a real, true, initial counting system. So let $$\{x\}\hookrightarrow X\rightarrow^{\!\!\!\!\!\!f}X$$ be any counting system. The set of all functions $$X\rightarrow X$$ is easily verified to be a monoid $\cal X$. Hence there is a unique monoid homomorphism $$N_0\rightarrow{\cal X}$$ sending $$1\mapsto f;$$ which is written $$n\mapsto f^n.$$ Now it is possible to define a mapping $$N_0\rightarrow X$$ by writing $$n\mapsto f^n(x)$$
- I assert that every number other than $1$ and $-1$ has indeed got a prime factor. Since the number $n$ in question is not a unit, the set of its multiples $${\frak a}=\{xn:x\in z\}$$ is not all of $Z$. Consider the class $\cal I$ of all proper ideals of $Z$ containing $\frak a$ as a subset; the set-inclusion relation $\subset$ makes $\cal I$ a partially ordered set. Now consider any subclass of $\cal I$ with the property that if $\frak b$, ${\frak c}\in{\cal C}$ then either ${\frak b}\subset{\frak c}$ or ${\frak c}\subset{\frak b}$; the union of $\cal C$ is trivially a proper ideal of $Z$ containing as a subset every ideal of $\cal C$ and also containing $\frak a$ as a subset. By Zorn's Lemma, a proper ideal $\frak m$ of $Z$ exists that is maximal with respect to the property of being a proper ideal of $Z$ containing $\frak a$ as a subset. Hence $\frak m$ is maximal with respect to the property of being a proper ideal; and hence is a prime ideal.
Now, in the ring $Z$ every ideal has a generator. The generator of a prime ideal is prime; since $n$ is in this ideal, we are done.