Show that $1439^2 | \sum_{k=1}^{1439}k^{1439}$

Question

I couldn't show that $1439^2 | \sum_{k=1}^{1439}k^{1439}$ But i showed that $1439 | \sum_{k=1}^{1439}k^{1439}$ maybe it can help.
We can easily prove that 1439 is a prime number .
According to Fermat's little theorem : $k^{1439}\equiv k\pmod{1439}$
Then , $\sum_{k=1}^{1439}k^{1439}\equiv \sum_{k=1}^{1439}k\pmod{1439}$
And it is commonly known that $\sum_{k=1}^{1439}k=\frac{1439\cdot 1440}{2}$
We can deduce that $\sum_{k=1}^{1439}k^{1439}\equiv 0 \pmod{1439}$
I hope so gonna help me out !

Answer 1

This is true for every odd number $n$, not just $1439$.

Indeed, pair the terms of the sum $S=\sum_{k=0}^{n} k^n$ as $a^n$ and $(n-a)^n$.

Now, $$(n-a)^n = n^n - n n^{n-1} a+ \binom{n}{2} n^{n-2} a^2 - \cdots - \binom{n}{2} n^2 a^{n-2} + n n a^{n-1} - a^n \equiv - a^n \bmod n^2 $$

Hence, $S \equiv 0 \bmod n^2$.