What is $15^{15} + 16^{16} + 17^{17} + 18^{18} + 19^{19} + 20^{20} \pmod{7}$? I am trying to evaluate $15^{15} + 16^{16} + 17^{17} + 18^{18} + 19^{19} + 20^{20} \pmod{7}$. I have found that $15^{15} \equiv 1 \pmod{7}$ and that $16^{16} \equiv 2 \pmod{7}$.
To evaluate $15^{15} \pmod{17}$, I did the following:
$$15 = 2 \times 7 + 1 \equiv 1 \pmod{7}$$
$$15^{15} \equiv 1 \pmod{7}$$
Then, to evaluate $16^{16}$, I wrote:
$$16 = 15 + 1 \equiv 1 + 1 = 2 \pmod{7}$$
$$16^{16} \equiv 2^{16} \pmod{7}$$
$$2^{3} = 8 = 7+1 \equiv 1 \pmod{7}$$
$$2^{16} = 2^{3} \times 2^{13} \equiv 2^{13} = 2^{3} \times 2^{10} \equiv 2^{10} \equiv \dots \equiv 2 \pmod{7}$$
Nevertheless, I have not managed to figure out how to evaluate $17^{17}$. How should I go about this and is my overall approach for evaluating the sum in question a good one?

Euler's Theorem states that
$$ a^{\phi(n)}\equiv1 \pmod{n}$$
For all $n$, where $\phi(n)$ is the Euler totient function, denoting the number of positive integers less than $n$ relatively prime to $n$. For primes, $\phi(n)=n-1$. Thus
$$ a^{n-1}\equiv1\pmod n. $$
This result is also known as Fermat's little theorem. Now, $7$ is prime, so anything to the power of $6$ is congruent to $1$. We have
$$\begin{split}
&15^{15}+16^{16}+17^{17}+18^{18}+19^{19}+20^{20}\\
\equiv\, & 15^3 + 16^4 + 17^5 + 18^0 + 19^1 + 20^2 \\
\equiv \, & 1^3 + 2^4 + 3^5 + 4^0 + 5^1 + 6^2\\
 \equiv\, & 1+16+243+1+5+36\\
\equiv\, & 302\\
\equiv\, & 1\pmod{7}.
\end{split}$$
