We need to determine all pairs $(n, p)$ of positive integers such that:
- $p$ is a prime,
- $n > 1$,
- $(p-1)^n + 1$ is divisible by $n^{p-1}$.

1. **Case: $p = 2$**

If $p = 2$, then we need $n \mid 1^n + 1 = 2$. Since $n > 1$, the only possible value for $n$ is $n = 2$. Therefore, $(n, p) = (2, 2)$ is a solution.

2. **Case: $p$ is odd**

Let $q$ be the smallest prime divisor of $n$ and let $o$ be the order of $p-1 \mod q$. Since $(p-1)^{2n} \equiv 1 \mod q$, we get $o \mid 2n$. By Fermat's Little Theorem, $o \mid q-1$.

If $o$ has a factor in common with $n$, this factor would be $\leq q-1 < q$, which is impossible. Therefore, $o \mid 2n$ reduces to $o \mid 2$, yielding $(p-1)^2 \equiv 1 \mod q$ or equivalently $q \mid p(p-2)$.

- **Subcase: $q \mid p-2$**

We get $(p-1)^n + 1 \equiv 1^n + 1 = 2 \mod q$. But we need $q$ to divide this number, thus $q = 2$. However, this implies $p$ is even, so $p = 2$, which we excluded in this case.

- **Subcase: $q \mid p$**

We write $n = p^k s$ where $p \nmid s$ and $k > 0$. It is well known that for odd primes $p$, if $v_p(a - b) > 0$ and $p \nmid a, b$, then $v_p(a^t - b^t) = v_p(a - b) + v_p(t)$ (where $v_p(m)$ is the $p$-adic valuation of $m$, the number of times $m$ is divisible by $p$). With $a = p-1$, $b = -1$, and $t = n$, this gives us $v_p((p-1)^n + 1) = k + 1$ and it has to be $\geq k(p-1)$ to enable $p^{k(p-1)} \mid n^{p-1} \mid (p-1)^{p^k s} + 1$. Now $k + 1 \geq k(p-1) \iff 2k + 1 \geq kp \iff 2 + \frac{1}{k} \geq p$. Since we assumed $p \geq 3$, we have $k = 1$ and $p = 3$.

If $s = 1$, we have $n = 3$ giving a solution. So let's assume $s > 1$. Let $q$ be the smallest prime divisor of $s$. By $n = 3$, we want $s \mid 9s^2 = n^2 \mid 2^n + 1 = 8^s + 1$. Following the same idea as before, we take $o$ as the order of $8 \mod q$ and get again $o \mid 2$ by the same argument. In both cases, $q \mid 8^2 - 1 = 63$, thus $q = 7$. But $8^s + 1 \equiv 1^s + 1 = 2 \mod 7$, which is a contradiction.

Conclusion: All solutions are given by $(n, p) = (2, 2), (3, 3)$.

The final answer is $\boxed{ (2, 2), (3, 3) }$.