In RSA, given public key $(n,e)$ and $d^e\bmod n$, can we factor $n$?
We can compute the amount of $E=e^{e-1} mod(n)$ then compute $C=c^E=(d^e)^{e^{e-1}}=d^{e^e} mod(n)$. Now we can compute the secret key $d$ as below:
$C^c=(d^{e^e})^{d^e} mod(n)=d^{e^e.d^e}=d^{(e.d)^e}=d^{1^e}=d \mod(n)$