What is the remainder if $$19^{20} - 20^{19}$$ is divided by 7?

Using Fermat's theorem :

If p is a prime number and a, p are co primes $$\left(a^{p-1}\right)\ mod\ p=1$$

Remainder when $$19^{20}$$ is divided by 7 = $$19^2$$ mod 7 =4. ( Here $$19^{20\ }=\ \left(\left(19\right)^6\right)^3\cdot\left(19\right)^2$$

Since the remainder for $$19^6$$ is 1 the remainder for $$\ 19^{20}$$ is  equivalent to the $$\frac{19^2}{7}$$ = 4.

Remainder when $$20^{19}$$ is divided by 7 = $$20^1$$ mod 7 =6.( Here $$\frac{20^6}{7}$$ the remainder is 1 and since 

$$20^{19}=\ \left(20^6\right)^3\cdot\left(20\right)^1\ =\ \frac{\left(1\cdot20\right)}{7}$$. The remainder is 6.

Remainder when $$19^{20} - 20^{19}$$ is divided by 7=4-6=-2=> 5.