Prove by induction that for $m, n ∈ N$, $m^{2n+1} − m$ is divisible by $6.$ 
Prove by induction that for $m, n ∈ N$, $m^{2n+1} − m$ is divisible by $6.$

What I have thus far:
Base case: $m=n=0$;
$0^{0+1}-0 = 0$, which is divisible by $6$.
Base case(2): $m=n=1$;
$1^{2+1}-1 = 0$, which is divisible by $6$.
Proposition: $m^{2n+1} − m = 6 \times k$ for $m,n ∈ N$.
Induction step: $m^{2n+1} − m$ => $m^{2(n+1)+1} -m$
And this is where I'm kinda stuck on.
I was also wondering if there maybe was an easier way to prove this by using $m^{2n+1} − m \mod 6 = 0$ as proposition.

Your induction step isn't right
$m^{2(n+1) + 1} - m = m^{2^n+1}m^2 - m=$
$([m^{2n+1} -m] + m)m^2 - m = $
$(6k + m)m^2 - m =$
$6km^2 + m^3- m$.
So we must prove $m^3 -m$ is always a multiple of $6$.
Hint:  $m^3 - m = m(m^2-1) = m(m-1)(m+1)$
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In general you can do two proofs by induction, one where one variable is is fixed and the other is run through in the proof.
That is to say.  Prove for a fixed $m$ that Base case $m^{2*0+1} - m$ is divisible by $6$; and induction step if $m^{2n+1} - m$ is divisible by $6$ then $m^{2(n+1) + 1} -m$ is divisible by $6$;  ANd then to a second indcution proof for a fixed $n$ that Base case $0^{2n+1} - 0$ is divisble by $6$; and induction step if $m^{2n+1} - m$ is divisble by $6$ then $(m+1)^{2n+1} -(m+1)$ is divisible by $6$.
In this case it's not so straight forward.  The induction on $n$ with $m$ fix will rely on us showing $m^3 - m$ is divisible but $6$.  And the induction on $m$ with $n$ fix will rely on showing $\sum_{k=1}^{2n} (-1)^k{2n+1\choose k}m^k$ is divisible by $6$.
But I think the easiest thing is to first prove that $m^2 -m$ is divisible by $6$.  This can be done by induction but it is easier to do it by modular arithmentic $\mod 2$ and $\mod 3$.
$m^3 -m = (m-1)m(m+1)$ so if $m\equiv 0,1,-1\pmod 3$ then $m,m-1,m+1 \equiv 0 \pmod 3$ so $3|(m-1)m(m+1)$ and if $m \equiv 0,1 \pmod 2$ then $m,m-1\equiv 0\pmod 2$ so $2|(m-1)m(m+1)$ and so $6|(m-1)m(m+1)$.
Then prove by induction for a fixed $m$ than induction on $n$ shows $6|m^{2m+1} - m$.
......
Or to use modular arithmetic from the start.
$m^{2m+1} - m = m(m^{2n} - 1)$.
If $m\not \equiv 0 \pmod{2,3}$ its easy to show therefore than $m^2\equiv 1\pmod {2,3}$ and therefore $m^{2n} \equiv 1 \pmod{2,3}$ and if $2,3 \not \mid m$ then $2,3 \mid m^{2n} -1$ so either way $6|m(m^{2n} -1)$
i.e.  either $m \equiv 0 \pmod 2$ or if $m\not \equiv 0\pmod 2$ then $m^2\equiv 1 \pmod 2$.
And either $m \equiv 0 \pmod 3$ or if $m\not \equiv 0 \pmod 3$ then $m^2 \equiv 1\pmod 3$.
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In hindsight, I think the easiest answer should have been: It's often easiest to prove things are divisible by $2$ and by $3$ separately.
It's easy to convince ourselves that $m$ and $m^{2n}$ must be both even or both odd so $m^{2n} -m$ must be even.  ($m\equiv 0, 1\pmod 2$ and $m^k\equiv 0^k, 1^k\equiv 0, 1\equiv m\pmod 2$ so $m^k-m \equiv m-m\equiv 0 \pmod 2$.) And it's not not much harder to convince ourselves that similarly that $m^2 \equiv \begin{cases} 0&m\equiv 0 \pmod 3\\1&m\equiv 1\pmod 3\\1&m\equiv 2\pmod 3\end{cases}\pmod 3$ so $m^{2k+1}=(m^2)^k m \equiv \begin{cases}0\cdot m =0\equiv m&m\equiv 0 \pmod 3\\1\cdot m \equiv m&\begin{cases}1\\2\end{cases}\pmod 3\end{cases}\equiv m \pmod 3$.  So $m^{2n+1} - m\equiv m-m\equiv 0 \pmod 3$.
And that would be all.
