A somewhat more interesting problem is as follows.
Show that $n-1$ divides $n^k - 1$ where $(n,k) \in \mathbb{N}$

It is not true in general that if $$ M^\prime \to M \to M^{\prime \prime}$$
is an exact sequence that the sequence
$$ M^\prime \otimes N \to M \otimes N \to M^{\prime \prime} \otimes N$$
is exact.

Without the FToA: Suppose $a,b,c,d$ are squarefree with $ab^2=cd^2$ and $(a,c)=(b,d)=1$. Then
$$b^2\mid cd^2,~ (b^2,d^2)=1\implies b^2|c\implies b,d=1.$$ Thus, losing the restriction $(a,c)=(b,d)=1$, we can apply the reasoning instead to $$\frac{a}{(a,c)},\frac{b}{(b,d)},\frac{c}{(c,d)},\frac{d}{(b,d)},$$ (which are also squarefree) thereby obtaining $b=d$ and $a=c$.