\begin{align} 0v &= (0+0)v \\ &= 0v+0v \end{align}

Given scalar multiplication is closed, \(0v \in V\), which means \(\exists -0v:0v+(-0v)=0\). Applying that to both sides:

\begin{equation} 0 = 0v\ \blacksquare \end{equation}

The opposite proof of \(\lambda 0=0\) but vectors work the same exact way.