Is there $n>1$ such that $n^{n+1} \equiv 1 \mod (n+1)^n$? First of all, note that $\frac{n^{n+1}}{(n+1)^n} \sim \frac{n}{e}$.  
Question: Is there $n>1$ such that $n^{n+1} \equiv 1 \mod (n+1)^n$?
There is an OEIS sequence for $n^{n+1}\mod (n+1)^n$: https://oeis.org/A176823. 

$0, 1, 8, 17, 399, 73, 44638, 1570497, 5077565, 486784401, 22187726197,
 166394893969, 13800864889148, 762517292682713, 9603465430859099,
 803800832678655745, 3180753925351614970, 947615093635545799201$

 A: We first prove that it is impossible when $n\not\equiv 1\pmod 4$.
Notice how:
$$n^{n+1}=1+(n-1)\sum_{k=0}^{n}n^k$$
So, we would have $n^{n+1}\equiv 1\pmod{(n+1)^n}$ if and only if:
$$(n+1)^n\mid (n-1)\sum_{k=0}^{n}n^k$$
And since the RHS won't be equal to $0$, we'd have:
$$(n-1)\sum_{k=0}^{n}n^k\ge(n+1)^n$$
but, since $4\nmid n-1$, we have $\gcd(n-1,(n+1)^n)\le 2$, so this becomes:
$$2\sum_{k=0}^{n}n^k\ge(n+1)^n$$
Multiplying both sides with $(n-1)$ yields:
$$2n^{n+1}>2(n^{n+1}-1)\ge (n-1)(n+1)^n=\frac{n-1}{n+1}\cdot(n+1)^{n+1}$$
or, assuming $n>1$:
$$\frac{2n+2}{n-1}>\left(\frac{n+1}{n}\right)^{n+1}=\left(1+\frac1n\right)^{n+1}>\left(1+\frac1n\right)^n$$
However, as $n$ tends to infinity, the LHS tends to $2$, while the RHS tends to $e$. Using induction, we first prove that for $n\ge 11$, we have $2.4\ge\frac{2n+2}{n-1}$. This is quite easy and I'll leave it out.
For $n=11$, we also have that the RHS is greater than $2.6$ and since $(1+\frac1n)^n$ keeps increasing as $n$ keeps increasing, this shows that there are no $n\not\equiv 1\pmod 4$ with $n\ge 11$ with $n^{n+1}\equiv1\pmod{(n+1)^n}$. Some quick testing reveals that there are no solutions at all for $n\not\equiv 1\pmod 4$.

As per @san's request, I'll also provide his solution for the case $n\equiv 1\pmod 4$, so that there is one complete answer. 
Assume by contradiction that $n=4j+1$ for some positive integer $j$, and  $n^{n+1} \equiv 1 \mod (n+1)^n$. Then
$$
n+1=2(2j+1)\quad\text{and}\quad n-1=2^ra
$$ 
for some $r\ge 2$ and some odd $a$.
There exists some $k$ such that $n^{n+1} - 1 =k\cdot (n+1)^n=k\cdot 2^n(2j+1)^n$.
But
\begin{eqnarray*}
n^{n+1} - 1&=&\sum_{s=0}^{n+1}\binom{n+1}{s}(n-1)^s-1\\
&=& \sum_{s=1}^{n+1}\binom{n+1}{s}(2^r a)^s\\
&=& (n+1)2^r a+2^{2r}a^2 \sum_{s=2}^{n+1}\binom{n+1}{s}(2^r a)^{s-2}\\
&=& (2j+1)2^{r+1}a+2^{2r}a^2 \sum_{s=2}^{n+1}\binom{n+1}{s}(2^r a)^{s-2}\\
&=& 2^{r+1}\left((2j+1)a+2^{r-1}a^2 \sum_{s=2}^{n+1}\binom{n+1}{s}(2^r a)^{s-2}\right)\\
\end{eqnarray*}
and $2^n$ divides $n^{n+1}-1$, hence $2^n$ divides $2^{r+1}$, and so $r\ge n-1$, which contradicts the fact that $n-1=2^ra$, since in general $n-1<2^{n-1}$.
A: Use the previous answer for the 
$n = 4k + 1$
$n^{n+1}-1 = k \dot (n-1)$
For the $n=2*k$ 
$n-1$ and $n+1$ is coprime => $n^{n+1}-1 = (n-1)(n+1)^n *k$. By the asymptotical equation from the comments k = 0.
For the $n=4k+3$.   $\frac{n-1}{n}$ and $n+1$  is coprime $n^{n+1}-1 = k* \frac{(n-1)(n+1)^n}{2} $. By the asymptotical equation from the comments k = 0.
