\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.