Prove that if $|x|<1$ then $x^{6}<1$ I am trying to prove that if $|x|&lt;1$ then $x^{6}&lt;1$ and that if $x^{6}&lt;1$ then $|x|&lt;1$.

For the first part, I thought of first considering $0&lt;x&lt;1$. Multiplying by $x$ (which is positive) I then get $0&lt;x^{2}&lt;x&lt;1$. And repeating the multiplication I would end up with $x^{6}&lt;1$.

Then I would consider $-1&lt;x&lt;0$. But now multiplying by $x$ (which is negative) gives $-x&lt;x^{2}$ and I dont see how to get $x^{2}&lt;1$ from here and then $x^{6}&lt;1$.
There is probably a more elegant way of going about this?

For the second part I thought of using the contrapositive and prove $x\leq -1$ or $x\geq 1$ then $x^{6}\geq 1$, and then I would try to use the previous proof to break this into two cases.

$x\leq -1$ then $x^{6}\geq 1$ and $x\geq 1$ then $x^{6}\geq 1$.
The argument that dxiv made in a comment is, I think, the most elegant argument.  As comments are ephemeral, I'll reiterate it here (with some additional details):  note that
$$ \frac{|x|^6 - 1}{|x|-1} = |x|^5 + |x|^4 + |x|^3 + |x|^2 + |x| + 1. \tag{1}$$
This is actually a specific case of a much more general result: for any natural number $n$,
$$ \frac{t^n-1}{t-1} = \sum_{j=0}^{n-1} t^j. $$
This can be proved by an induction argument after noticing that
$$t^{n+1} - 1 = t^{n+1} - t^n + t^n -1 = t^n(t-1) + t^n - 1.$$  In this particular case, we take $t = |x|$.  In any event, since all of the terms on the right-hand side of (1) are nonnegative (and $1&gt;0$, i.e. $1$ is strictly positive), it follows that
$$ \frac{|x|^6 - 1}{|x|-1} &gt; 0. \tag{2}$$
Note that
$$ \frac{a}{b} &gt; 0 \iff (a&gt;0 \land b &gt; 0) \lor (a&lt; 0\land b &lt; 0). $$
That is, a fraction is positive if and only if both the numerator and denominator have the same sign.  Applying this to (2), either
$$ |x|^6 - 1 &lt; 0 \qquad\text{and}\qquad |x| - 1 &lt; 0, \tag{3} $$
or
$$ |x|^6 - 1 &gt; 0 \qquad\text{and}\qquad |x| - 1 &gt; 0. $$
Therefore if we assume that $|x|&lt;1$ it immediately follows from (3) that $|x|^6&lt;1$, and vice versa.

Alternatively, we can make your approach work.  The first part of your argument is basically an induction argument.  We can actually strengthen that result a bit and get the following:

Proposition: If $|x| &lt; 1$ then $|x|^n &lt; 1$ for any natural number $n$.

Proof:  The proof is by induction.  Assume that $|x|&lt; 1$, and note that $|x|^1 &lt; 1$, providing a base for the induction.  Now suppose that $|x|^k &lt; 1$.   Then
$$ |x|^k &lt; 1 \implies |x|^k \cdot |x| &lt; 1 \cdot |x| \implies |x|^{k+1} &lt; |x|. $$
But $|x|&lt;1$, which proves the result.$\tag*{$\blacksquare$}$
Taking $n = 6$ gives gives the result in the original question.  Then, as you note, the converse requires an additional argument.  Following your instinct of attempting contraposition, you might consider the following proposition:

Proposition: Let $n$ be a natural number.

If $|x| &lt; 1$ then $|x|^n &lt; 1$,
if $|x| = 1$ then $|x|^n = 1$, and
if $|x| &gt; 1$ then $|x|^n &gt; 1$.


There may be more elegant arguments, but the induction proof above can be used, mutatis mutandis, to prove the two new statements.  This gives a kind of trichotomy, from which you can conclude that if $|x|^n &lt; 1$, then $|x| &lt; 1$.
suppose $0&lt;x&lt;1$ and $x^6\ge1$

$x^6=x\cdot x^5$ and since $0&lt;x&lt;1$ multiplying by $x$ is like scaling a value making it $100\cdot x$% of the original value, which is smaller than $&lt;1\cdot100$%.

So $x^6=x\cdot x^5&lt;x^5&lt;...&lt;x\cdot x&lt;x&lt;1$ and that's impossible