Does $n^n\mod p$ repeat every $2p$ terms? Going through some random numbers, I have found that $n^n\mod10$ repeats every $20$ terms, and it more or less is as follows.
$$1^1\equiv1\pmod{10}$$
$$2^2\equiv4\pmod{10}$$
and so on.  The sequence I found was $1,4,7,6,5,6,3,6,9,0,1,6,3,6,5,6,7,4,9,0,\dots$  After that, it appears to repeat.
In modular $3$: $1,1,0,1,2,0,\dots$ and it appears to repeat as well.
Is there an explanation for this?  And is there anything provable about this?
Naturally, I don't mind numerical results that disprove my statement, but even so, it is interesting that the numbers appear to repeat.
 A: This is not quite true, but it's very close to being true. In particular, the truth of this hinges on the following statement:

The sequence $1,n,n^2,n^3,\ldots$ is eventually periodic mod any $m$.

More strongly, one can find that the period of this sequence will always divide $\lambda(m)$ where $\lambda$ is the Carmichael function, which is the smallest $k$ such that $n^k\equiv 1\pmod m$ for any $n$ coprime to $m$. This is almost tautological - the only step of interest is that if $n$ isn't coprime to $m$, one can take the largest divisor $m'$ of $m$ such that $m'$ and $n$ are coprime, and see that the period of the sequence divides $\lambda(m')$ which divides $\lambda(m)$.
In particular, this tells us that the sequence $n^n$ mod $m$ will be eventually periodic with a period of $P=\operatorname{lcm}(m,\lambda(m))$. We may prove this since we can first get that, since $m$ divides $P$, we have
$$n^{n}\equiv (n+P)^n\pmod m$$
and then since, for large enough $n$ we have that $x^n=x^{n+\lambda(m)}$ for any $x$, we get that $x^n=x^{n+P}$ since $\lambda(m)$ divides $P$, so:
$$n^n\equiv (n+P)^{n+P}\pmod m$$
as desired. It seems reasonable that $P$ is the fundamental period of the sequence $n^n$, but I haven't managed to prove this yet. (It is, however, possible to see that the lcm of the fundamental period and $m$ is $P$, but this doesn't quite suffice).
In particular, the quantity $\operatorname{lcm}(m,\lambda(m))$ is $20$ when $m=10$ and is $6$ when $m=3$. However, when $m=5$, this quantity is $20$ and the sequence is only periodic with period $20$. The first $40$ terms are as follows:
\begin{align*}&1, 4, 2, 1, 0, 1, 3, 1, 4, 0, 1, 1, 3, 1, 0, 1, 2, 4, 4, 0,\\ &1, 4, 2,
1, 0, 1, 3, 1, 4, 0, 1, 1, 3, 1, 0, 1, 2, 4, 4, 0,\ldots\end{align*}
