proving the laplacian of a vector in cylindrical coordnates

I am proving the following identity for the laplacian of a vector $\vec{v}=<v_r,v_\theta,v_z>$ in cylindrical coordinates: $$\nabla^2 \vec{v}=\left( \frac{\partial^2 v_r}{\partial r^2}+\frac{1}{r^2}\frac{\partial^2 v_r}{\partial \theta^2}+\frac{\partial^2 v_r}{\partial z^2}+\frac{1}{r}\frac{\partial v_r}{\partial r}-\frac{2}{r^2}\frac{\partial v_\theta}{\partial \theta} -\frac{v_r}{r^2}\right )\vec{e_r} \\ + \left (\frac{\partial^2 v_\theta}{\partial r^2}+\frac{1}{r^2}\frac{\partial^2 v_\theta}{\partial \theta^2}+\frac{\partial^2 v_\theta}{\partial z^2}+\frac{1}{r}\frac{\partial v_\theta}{\partial r}+\frac{2}{r^2}\frac{\partial v_r}{\partial \theta}-\frac{v_\theta}{r^2} \right )\vec{e_\theta} \\ \left( \frac{\partial^2 v_z}{\partial r^2}+\frac{1}{r^2}\frac{\partial^2 v_z}{\partial \theta^2}+\frac{1}{r}\frac{\partial v_z}{\partial r}+\frac{\partial^2 v_z}{\partial z^2} \right)\vec{e_z}$$ I am able to derive the following identity for the Laplacian operator in cylindrical coordinates $$\nabla^2=\frac{\partial^2}{\partial r^2}+\frac{1}{r}\frac{\partial}{\partial r}+\frac{1}{r^2}\frac{\partial^2}{\partial \theta^2}+\frac{\partial^2}{z^2} $$. So to prove the desired identity, $$\nabla^2 \vec{v}=\left(\frac{\partial^2}{\partial r^2}+\frac{1}{r}\frac{\partial}{\partial r}+\frac{1}{r^2}\frac{\partial^2}{\partial \theta^2}+\frac{\partial^2}{z^2} \right)(v_r\vec{e_r}+v_\theta \vec{e_\theta}+v_z\vec{e_z}) \\ = \left(\frac{\partial^2}{\partial r^2}+\frac{1}{r}\frac{\partial}{\partial r}+\frac{1}{r^2}\frac{\partial^2}{\partial \theta^2}+\frac{\partial^2}{z^2} \right)(v_r\vec{e_r})+ \left(\frac{\partial^2}{\partial r^2}+\frac{1}{r}\frac{\partial}{\partial r}+\frac{1}{r^2}\frac{\partial^2}{\partial \theta^2}+\frac{\partial^2}{z^2} \right)(v_\theta\vec{e_\theta})+ \left(\frac{\partial^2}{\partial r^2}+\frac{1}{r}\frac{\partial}{\partial r}+\frac{1}{r^2}\frac{\partial^2}{\partial \theta^2}+\frac{\partial^2}{z^2} \right)(v_z\vec{e_z})$$. And upon distributing the vector components to the operator I finally get $$\nabla^2 \vec{v}=\left( \frac{\partial^2 v_r}{\partial r^2}+\frac{1}{r}\frac{\partial v_r}{\partial r}+\frac{1}{r^2}\frac{\partial^2 v_r}{\partial \theta^2}+\frac{\partial^2 v_r}{\partial z^2}-\frac{v_r}{r^2} \right)\vec{e_r} \\ +\left( \frac{\partial^2 v_\theta}{\partial r^2}+\frac{1}{r}\frac{\partial v_\theta}{\partial r}+\frac{1}{r^2}\frac{\partial^2 v_\theta}{\partial \theta^2}-\frac{v_\theta}{r^2}+\frac{\partial^2 v_\theta}{\partial z^2} \right)\vec{e_\theta} \\ \left( \frac{\partial^2 v_z}{\partial r^2}+\frac{1}{r}\frac{\partial v_z}{\partial r}+\frac{1}{r^2}\frac{\partial^2 v_z}{\partial \theta^2}+\frac{\partial^2 v_z}{\partial z^2} \right)\vec{e_z}$$ which is not the same with the identity. I am confused how the $-\frac{2}{r^2}\frac{\partial v_\theta}{\partial \theta}$ and $\frac{2}{r^2}\frac{\partial v_r}{\partial \theta}$ appeared in the $\vec{e_r}$ and $\vec{e_\theta}$ components, respectively. Where did I go wrong? Need help...thanks

The product rule for second order differentiation is $(fg)'' = f''g + 2f'g'+ fg''$. You simply omitted the middle value.

In Cartesian coordinates, the Laplacian of a vector can be found by simply finding the Laplacian of each component, $\nabla^{2} \mathbf{v}=\left(\nabla^{2} v_{x}, \nabla^{2} v_{y}, \nabla^{2} v_{z}\right)$. However, as noted above, in curvilinear coordinates the basis vectors are in general no longer constant but vary from point to point. You will need either to derive these vectors or use the general definition of the Laplacian of a vector,

$$\nabla^{2} \mathbf{v}=\nabla(\nabla \cdot \mathbf{v})-\nabla \times(\nabla \times \mathbf{v})$$