How to reduce congruence power modulo prime? If I have a congruence equation, says 
$$x^{15} - x^{10} + 4x - 3 \equiv 0 \pmod{7}$$
Then can I use Fermat's little theorem like this:
$$(x^{6})^2 \cdot x^3 - x^6 \cdot x^4 + 4x - 3 \equiv 0 \pmod{7}$$
$$ x^3 - x^4 + 4x - 3 \equiv 0 \pmod{7}$$
Update
Should it be 
$$x^{14}x - x^7x^3 - 4x - 3 \equiv 0 \pmod{7}$$
$$x^2x - x.x^3 - 4x - 3 \equiv 0 \pmod{7}$$
$$x^3 - x^4 - 4x - 3 \equiv 0 \pmod{7}$$
?
Thanks, 

Not quite. Look for example at the congruence $x^6 \equiv 0 \pmod{7}$.  If one assumes that $x^6 \equiv 1$, things go bad.  In this case it is easy to spot that there is a problem, but perhaps in a more complicated setting one might miss it.
I would advise using the fact that $x^7 \equiv x \pmod{7}$, basically a variant of Fermat's Theorem that holds always, not just almost always.
