Take a basis $\\{v_1,...,v_k \\} $ of $V$, and define projection $\pi_i \in V^* $ as $$ \pi_i(\sum_j c_j v_j) = c_i $$ Then $\\{\pi_1,...,\pi_k \\} $ is also a basis of $V^*$ . Now introduce the $l^1$ norm as $ \|v\| = \sum_i |\pi_i(v)| $ for all $ v \in V $. Now if $v^n \rightharpoonup v \in V $ then $ \pi_i(v^n) \rightarrow \pi_i(v) $ for all $i$, hence $ \|v^n -v\| \rightarrow 0 $. But in finite dimensional space, all norms are equivalent, hence the convergence is strong.