Solution : Either I am super drunk or the problem is too easy . Suppose $$n^{n-p}=m^n$$for some $m \in N.$ Now observe that since $p>n$ we have $n-p<0$ so $$m^n=n^{n-p} \neq integer$$unless $n=1$. So $n=1$ is the only possible case and hence $(n,p)=(1,p)$ where $p$ is any prime . In that case $(4,2)$ works, and to look whether is there any other solutions we have to consider prime factorisation of $n$ and what happens if $p$ does not divide $n$.Will add the full solution later.

Fun

The condition is $n>p$, sorry for the mistake.

If someone wants a short hint , Check what happens if p doesnt divide n .

I claim $(n,p)=(4,2)$ is the only solution. Let $n=p_1^{e_1}\cdots p_L^{e_L}$ for primes $p_1<\cdots<p_L$ and $e_i\ge 1$. Note that $n\mid e_i(n-p)$ for every $i$. If $p\nmid n$, then $(n-p,n)=1$, so $n\mid e_i,\forall i$. This yields $n\ge 2^n$, which is absurd. So, $p\mid n$, set $n=pk$ with $k\ge 2$. We have that for some $u$, $(pk)^{pk-p} = u^{pk}$, that is $(pk)^{k-1}=u^k$. From here, it is evident that $pk$ is a perfect $k{\rm th}$ power, so set $pk=\ell^k$. We have in particular $p\mid \ell$, which, in turn, yields $p^{k-1}\mid k$. So, $k\ge p^{k-1}$. If $p\ge 3$, then we trivially have $k\ge (1+2)^{k-1}\ge 1+2(k-1)$, which is false for $k>1$. So, $p=2$ and $k\ge 2^{k-1}$. We easily see that the only solution to this inequality is $k\in\{1,2\}$; for $k=2$ we obtain $(n,p)=(4,2)$ as claimed.