Define $n$th term $a(n)=a(n-1)^2+a(n-1)$ and $a(0)=1$ sequence a(n)
$a(n)=1,2,6,42,1806,3263442,...$
Define $S(x,y)=1^y+2^y+3^y+...+x^y$
Can it be shown that
For all $k\le n$ $$S(a(k),a(n-1))\equiv1\pmod{a(k)}$$
For example let $n=2$
$a(0)=1,\ \ a(1)=a(n-1)=2,\ \ a(2)=a(n)=6$
$$\begin{split} S(1,2)&=1&\equiv 1\pmod1\\ S(2,2)&=5&\equiv1\pmod2\\S(6,2)&=91&\equiv1\pmod6\end{split}$$
I checked $n$ upto $4$
Does not work for $n=5$ and $k=5$, and gives $S(a(5),a(4))= 1807 \mod a(5)$
Strange enough $1807 = a(4)+1$, so it could be that there is some sort of connection.