How do I show that $u(x, t) = e^{-\alpha^2k^2t}\sin(kx)$ is a solution for $u_t = \alpha^2u_{xx}$.

Question

I want to show that $u(x, t) = e^{-\alpha^2k^2t}\sin(kx)$ is a solution for $u_t = \alpha^2u_{xx}$.

I did the following:

$$u_x = e^{-\alpha^2k^2t}\cos(kx)k$$

$$u_{xx} = -e^{-\alpha^2k^2t}\sin(kx)k^2$$

$$u_{t} = -\alpha^2k^2e^{-\alpha^2k^2t}\sin(kx)$$

so, plugging $u_{xx}$ into the equation $u_t = \alpha^2u_{xx}$ gives the following:

$$u_t = \alpha^2[-e^{-\alpha^2k^2t}\sin(kx)k^2]$$

after rearranging, we get:

$$u_{t} = -\alpha^2k^2e^{-\alpha^2k^2t}\sin(kx)$$

Is this all I have to do?