Show that $1^7+7^7+13^7+19^7+23^7\equiv{0}\pmod{63}$ Show that $1^7+7^7+13^7+19^7+23^7\equiv{0}\pmod{63}$
According to Fermat's theorem:
$$1^7+7^7+13^7+19^7+23^7\equiv{1+7+13+19+23}\pmod{7}\equiv{63}\pmod{7}\equiv{0}\pmod{7}$$
Now we need to show: $1^7+7^7+13^7+19^7+23^7\equiv{0}\pmod{9}$ , but how??

Since OP demands a proof without Euler's theorem, below is one:
First we denote the sum as $S$ and modulo $9:$
$$S\equiv1^7+(-2)^7+4^7+1^7+5^7\pmod9.$$
Since $(-2)^3\equiv1\pmod9,$ we find that $(-2)^7\equiv-2\pmod9,$ so that $$S\equiv4^7+5^7\pmod9.$$
But $5\equiv-4\pmod9,$ so $$S\equiv4^7+(-4)^7\equiv0\pmod9.$$  
Hope this helps.
