Our hypothesis and our conclusion use the term linear independence, so it will get a workout. To establish linear independence, we begin with the definition (Definition LICV) and write a relation of linear dependence (Definition RLDCV),
\begin{equation*}
\alpha_1\left(\vect{v}_1\right)+
\alpha_2\left(\vect{v}_1+\vect{v}_2\right)+
\alpha_3\left(\vect{v}_1+\vect{v}_2+\vect{v}_3\right)+
\alpha_4\left(\vect{v}_1+\vect{v}_2+\vect{v}_3+\vect{v}_4\right)
=\zerovector
\end{equation*}
Using the distributive and commutative properties of vector addition and scalar multiplication (Theorem VSPCV) this equation can be rearranged as
\begin{equation*}
\left(\alpha_1+\alpha_2+\alpha_3+\alpha_4\right)\vect{v}_1+
\left(\alpha_2+\alpha_3+\alpha_4\right)\vect{v}_2+
\left(\alpha_3+\alpha_4\right)\vect{v}_3+
\left(\alpha_4\right)\vect{v}_4
=
\zerovector
\end{equation*}
However, this is a relation of linear dependence (Definition RLDCV) on a linearly independent set, $\set{\vect{v}_1,\,\vect{v}_2,\,\vect{v}_3,\,\vect{v}_4}$ (this was our lone hypothesis). By the definition of linear independence (Definition LICV) the scalars must all be zero. This is the homogeneous system of equations,
\begin{align*}
\alpha_1+\alpha_2+\alpha_3+\alpha_4&=0\\
\alpha_2+\alpha_3+\alpha_4&=0\\
\alpha_3+\alpha_4&=0\\
\alpha_4&=0
\end{align*}
Row-reducing the coefficient matrix of this system (or backsolving) gives the conclusion
\begin{align*}
\alpha_1=0&&\alpha_2=0&&\alpha_3=0&&\alpha_4=0
\end{align*}
This means, by Definition LICV, that the original set
\begin{equation*}
\set{
\vect{v}_1,\,
\vect{v}_1+\vect{v}_2,\,
\vect{v}_1+\vect{v}_2+\vect{v}_3,\,
\vect{v}_1+\vect{v}_2+\vect{v}_3+\vect{v}_4
}
\end{equation*}
is linearly independent.