Prove that if $u(x,y)$ is harmonic then there is a $v(x,y)$ such that $f(z) = u + iv$ is analytic.

Question

The problem has the following hint: show that $F = (-u_y, u_x)$ is a gradient field.

So I took the hint and used $\nabla \times F = \hat{k}(u_{xx} + u_{yy})=0 $

Therefore there exists some function $f(x,y)$ such that $F = \nabla f(x,y) $, then:

$(f_x, f_y) = (-u_y, u_x)$ , which are two equations:

(1) $f_x = -u_y$, which after using Cauchy Riemann to force the condition becomes: $f_x = v_x$

(2) $f_y = u_x$ which after the same procedure becomes: $f_y = v_y$.

So then $f = v$

And this is where I am stuck... how can I prove that $u+iv$ is analytic?