linear algebra - Vector spaces - Multiplying by $-1$ yields inverse element of vector addition.

Please rate and comment. I want to improve; constructive criticism is highly appreciated.
Please take style into account as well.

The following proof is based on vector space related axioms.
Axiom names are italicised.
They are defined in Wikipedia (see vector space article).
Additionally, this trivial result is used.

Multiplying by $-1$ yields inverse element of vector addition.

Let $V$ be a vector space over a field $F$.
\begin{array}{lrll}
\text{By} & \dots & \text{we denote} & \dots \\
\hline
& 1 && \text{a multiplicative identity element in $F$.} \\
& (-1) && \text{an additive inverse of $1$ in $F$.} \\
& 0 && \text{an additive identity element in $F$.} \\
& \mathbf{0} && \text{an additive identity element in $V$.} \\
\end{array}

Let $\mathbf{v} \in V$.
We want to prove that
$$(-1)\mathbf{v} \text{ is an additive inverse of } \mathbf{v} \text{ in } V.$$
Proof. We prove that $\mathbf{v} + (-1)\mathbf{v} = \mathbf{0}$.
\begin{align*}
\mathbf{v} + (-1)\mathbf{v}
&= 1\mathbf{v} + (-1)\mathbf{v} && \text{by }\textit{Identity element of scalar multiplication} \\
&= (1 + (-1))\mathbf{v} && \text{by }\textit{Distrib. of scalar mult. (field addition)} \\
&= 0\mathbf{v} && \text{by }\textit{Inverse elements of field addition} \\
&= \mathbf{0} && \text{by the result linked above}

\end{align*}
QED