Is this a correct way to do this proof? Prove that if $\{\vec{v}_1,\dots,\vec{v}_k\}$ is linearly independent, then $\{\vec{v}_1,\dots{},\vec{v}_{k-1}\}$ is also linearly independent.
My proof:
Since $\{\vec{v}_1,\dots,\vec{v}_k\}$ is linearly independent, then the equation $$c_1\vec{v}_1+\cdots+c_k\vec{v}_k=\vec{0}$$ has only the trivial solution. This then implies that $$c_1\vec{v}_1+\cdots+c_{k-1}\vec{v}_{k-1}=-c_k\vec{v}_k$$ also has only the trivial solution by basic algebra. $-c_k\vec{v}_k=0$ so $\{\vec v_1, \ldots,\vec{v}_{k-1}\}$ is also linearly independent.
Is this proof correct?

If we suppose that $\{\vec{v}_1,\ldots,\vec{v}_{k-1}\}$ is linearly dependent then there's $c_1,\ldots,c_{k-1}$ not all zero such that 
$$c_1\vec{v}_1+\cdots+c_{k-1}\vec{v}_{k-1}=0$$
then take $c_k=0$ and we find
$$c_1\vec{v}_1+\cdots+c_k\vec{v}_k=\vec{0}$$
and this a contradiction with the fact that $\{\vec{v}_1,\ldots,\vec{v}_{k}\}$ is linearly independent.
