The result I am trying to prove:
$$n-m \Biggl\lfloor \frac{n}{m} \Biggr\rfloor =1 \Leftarrow\Rightarrow n^k\equiv 1 (\operatorname{mod} m)\quad \forall k \in \mathbb N \land n,m \in \mathbb N \,\backslash\, {\{1}\}$$
$$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad(0.0)$$
My attempt thus far:
$$n-m \Biggl\lfloor \frac{n}{m} \Biggr\rfloor =1 \Leftarrow\Rightarrow n-1-m \Biggl\lfloor \frac{n-1}{m} \Biggr\rfloor =0 $$
$$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad(1.0)$$
$$n-1-m \Biggl\lfloor \frac{n-1}{m} \Biggr\rfloor =0 \Leftarrow\Rightarrow n\equiv 1 (\operatorname{mod} m)$$
$$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad(2.0)$$
Using the compatibility of the congruence relation with multiplication:
$$a_1\equiv a_2 (\operatorname{mod} N) \land b_1\equiv b_2 (\operatorname{mod} N) \Rightarrow a_1b_1\equiv a_2 b_2 (\operatorname{mod} N) $$
$$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad(3.0)$$
By simply substituting $a_1=b_1=n$, $a_2=b_2=1$ and $N=m$:
$$n\equiv 1 (\operatorname{mod} m)\Rightarrow n^2\equiv 1 (\operatorname{mod} m) $$
$$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad(4.0)$$
and repeating this substitution a total number of times equal to $2\Bigl\lfloor\frac{k}{2} \Bigr\rfloor-\frac{1}{2}((-1)^k-1)$ recursively will yield:
$$n^{2\Bigl\lfloor\frac{k}{2} \Bigr\rfloor-\frac{1}{2}((-1)^k-1)}\equiv 1 (\operatorname{mod} m) \Rightarrow n^k\equiv 1 (\operatorname{mod} m) $$
$$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad(5.0)$$
And this inductively proves the result for all naturals $k$.
edit* I think I have missed a lemma of establishing $n$ and $m$ are coprime