Struggling with a proof that $-m=(-1)m$ Prove that

For all integers $m$, $-m=(-1)m$.
  Any help would be greatly appreciated.

 A: The integer $-m$ is the opposite of $m$ that's the integer such that
$$(-m)+m=0$$
and notice that the opposite is unique (the proof is pretty easy).
Now to prove the desired result:
$$(-1)m+m=\left[(-1)+1\right]m=0m=0$$
where we used  the distribituve law and the fact that
$$0\times a=0\quad \forall a$$
so by unicity 
$$-m=(-1)m$$
A: Lemma: for all $m$, $$0\cdot m=0$$
Proof: $$\begin{align}0\cdot m+0\cdot m&=(0+0)\cdot m&&\mbox{by the distributive axiom}\\
&=0\cdot m &&\mbox{by the defining property of $0$}\\
(0\cdot m+0\cdot m)+-(0\cdot m)&=0\cdot m+-(0\cdot m)&&\mbox{adding the same quantity to both sides}\\
0\cdot m+(0\cdot m+-(0\cdot m))&=0\cdot m+-(0\cdot m)&&\mbox{associativity of addition}\\
0\cdot m+0&=0&&\mbox{defining property of negatives}\\
0\cdot m&=0&&\mbox{defining property of $0$}\\
\end{align}$$

Now $$\begin{align}
(-1)\cdot m + m &=(-1)\cdot m + 1\cdot m &&\mbox{by the defining property of $1$}\\
&=(-1+1)\cdot m&&\mbox{by the distributive axiom}\\
&=0\cdot m&&\mbox{defining property of negatives}\\
&=0&&\mbox{by the lemma}\\
((-1)\cdot m + m)+-m&=0+-m&&\mbox{adding the same quantity to both sides}\\
(-1)\cdot m +( m+-m)&=0+-m&&\mbox{associativity of addition}\\
(-1)\cdot m +0&=0+-m&&\mbox{defining property of negatives}\\
(-1)\cdot m &=-m&&\mbox{defining property of $0$}\\
\end{align}$$
