Yes! This can be shown using the isomorphism $H^1(X,\mathcal{O}_X) \cong \mathrm{Pic} X$. First, look at the short exact sequence:

$ 0 \to \mathcal{O}_X^* \to \bigoplus \mathcal{O}_{X_i}^* \to \mathcal{C} \to 0$

From the long exact sequence of cohomological groups associated to the short exact sequence, it suffices to show that $H^1(X,\mathcal{C})\cong 0$. However, this is clear since from the short exact sequence above, we can see that the support of $\mathcal{C}$ is a finite number of points (points that belong to more than just 1 irreducible component) and hence, of dimension 0. Now, use Grothendieck's vanishing theorem and we are done.