Proof (1) Since $(p-1)^{p}+1$
$$\begin{array}{l} =(p-1+1)\left[(p-1)^{p-1}-(p-1)^{p-2}+(p-1)^{p-3}-\cdots\right. \\ \left.+(p-1)^{2}-(p-1)+1\right] \\ = p \sum_{i=0}^{p-1}(-1)^{i}(p-1)^{i} \\ = p\left[1+(p-2) \sum_{j=1}^{p-1}(p-1)^{2 j-1}\right] \\ > p\left(1+2 \times \frac{p-1}{2}\right)=p^{2}, \\ \text { and } \sum_{i=0}^{p-1}(-1)^{i}(p-1)^{i} \equiv \sum_{i=0}^{p-1}(1-i p) \equiv p-p \sum_{i=0}^{p-1} i \\ \equiv p-p \times \frac{p(p-1)}{2} \equiv p\left(\bmod p^{2}\right), \end{array}$$

Therefore, $(p-1)^{p}+1$ contains a prime factor different from $p$.
(2) Suppose $q(\neq p)$ is another prime factor of $(p-1)^{p}+1$, it is easy to see that $q \neq 2$, then $(p-1)^{2 p} \equiv 1(\bmod q)$.
Since $q \mid\left[(p-1)^{p}+1\right]$, we have $(p-1, q)=1$.
By Fermat's Little Theorem, we get $(p-1)^{q-1} \equiv 1(\bmod q)$.
Let $(q-1,2 p)=d$, then from $(p-1)^{2 p} \equiv 1(\bmod q),(p-1)^{q-1} \equiv 1(\bmod q)$, we get $(p-1)^{d} \equiv 1(\bmod q)$.
This is because there must exist a positive integer $s$, satisfying
$$s=\min \left\{x \in \mathbf{Z}_{+} \mid(p-1)^{x} \equiv 1(\bmod q)\right\} \text {. }$$

Let $2 p=a s+b(0 \leqslant b \leqslant s-1)$, then by $(p-1)^{2 p} \equiv(p-1)^{a s+b} \equiv(p-1)^{b}(\bmod q)$ and the definition of $s$, we know $b=0$, i.e., $s \mid 2 p$.
Similarly, $s \mid(q-1)$.
Thus, $s \mid(2 p, q-1)$.
Hence, $(p-1)^{d} \equiv 1(\bmod q)$.
Since
$(q-1,2 p)=d$, $d$ is $1, 2, p$ or $2 p$.
(i) If $d=1$ or $p$, then $(p-1)^{p} \equiv 1(\bmod q)$, which contradicts $q \mid\left[(p-1)^{p}+1\right]$.
(ii) If $d=2$, then $(p-1)^{2} \equiv 1(\bmod q)$, hence
$(p-1)^{p-1} \equiv 1(\bmod q)$,
$(p-1)^{p} \equiv(p-1)(\bmod q)$.
From this, we know $(p-1)^{p}+1 \equiv p(\bmod q)$, a contradiction.
Therefore, it must be that $d=2 p$.
Thus, $2 p \mid(q-1)$, which implies $q>p$.
Let the prime factors of $(p-1)^{p}+1$ be $p_{i}(i=1,2, \cdots, n)$.
Let $\beta_{i}=\alpha_{i} \log _{p} p_{i}$, then $p_{i}^{\alpha_{i}}=p^{\beta_{i}}$.
Since the function $x \mapsto \frac{x}{\ln x}$ is monotonically increasing on $[e,+\infty)$, we have
$\alpha_{i} p_{i}=\beta_{i} \ln p \cdot \frac{p_{i}}{\ln p_{i}} \geqslant \beta_{i} \ln p \frac{p}{\ln p}=\beta_{i} p$.
Therefore, $\sum_{i=1}^{n} \alpha_{i} p_{i} \geqslant p \sum_{i=1}^{n} \beta_{i}$.
Moreover, since
$$\begin{array}{l} \sum_{i=1}^{n} \beta_{i}=\sum_{i=1}^{n} \alpha_{i} \log _{p} p_{i}=\log _{p}\left[(p-1)^{p}+1\right] \geqslant p \log _{p}(p-1) \geqslant \frac{p}{2}, \\ \text { hence } \sum_{i=1}^{n} \alpha_{i} p_{i} \geqslant p \sum_{i=1}^{n} \beta_{i} \geqslant \frac{p^{2}}{2} . \end{array}$$