Convert the point $(\sqrt{2},-\sqrt{2})$ in rectangular coordinates to polar coordinates.  Enter your answer in the form $(r,\theta),$ where $r > 0$ and $0 \le \theta < 2 \pi.$
We have that $r = \sqrt{(\sqrt{2})^2 + (-\sqrt{2})^2} = 2.$  Also, if we draw the line connecting the origin and $(\sqrt{2},-\sqrt{2}),$ this line makes an angle of $\frac{7 \pi}{4}$ with the positive $x$-axis.

[asy]
unitsize(0.8 cm);

draw((-2.5,0)--(2.5,0));
draw((0,-2.5)--(0,2.5));
draw(arc((0,0),2,0,315),red,Arrow(6));
draw((0,0)--(sqrt(2),-sqrt(2)));

dot((sqrt(2),-sqrt(2)), red);
label("$(\sqrt{2},-\sqrt{2})$", (sqrt(2),-sqrt(2)), NE, UnFill);
dot((2,0), red);
[/asy]

Therefore, the polar coordinates are $\boxed{\left( 2, \frac{7 \pi}{4} \right)}.$
We have that $r = \sqrt{(\sqrt{2})^2 + (-\sqrt{2})^2} = 2.$  Also, if we draw the line connecting the origin and $(\sqrt{2},-\sqrt{2}),$ this line makes an angle of $\frac{7 \pi}{4}$ with the positive $x$-axis.

[asy]
unitsize(0.8 cm);

draw((-2.5,0)--(2.5,0));
draw((0,-2.5)--(0,2.5));
draw(arc((0,0),2,0,315),red,Arrow(6));
draw((0,0)--(sqrt(2),-sqrt(2)));

dot((sqrt(2),-sqrt(2)), red);
label("$(\sqrt{2},-\sqrt{2})$", (sqrt(2),-sqrt(2)), NE, UnFill);
dot((2,0), red);
[/asy]

Therefore, the polar coordinates are $\boxed{\left( 2, \frac{7 \pi}{4} \right)}.$
The final answer is $$\boxed{\left( 2, \frac{7 \pi}{4} \right)}$. I hope it is correct.