Show that this $\sum_{i=0}^{p-1}(-1)^i\binom{p^2-p}{pi}\equiv p^{p-1}\pmod {p^p}$ Let $p$ is an odd prime,show that
$$\sum_{i=0}^{p-1}(-1)^i\binom{p^2-p}{pi}\equiv p^{p-1}\pmod {p^p}$$
 A: Define $\zeta=e^{2\pi i/p}\in\mathbb{C}$, a primitive $p$-th root of $1$. We can use the identity
$$
\sum_{n=0}^{p-1}\zeta^{kn}=\begin{cases}p&:p|k,\\0&:p\nmid k,\end{cases}
$$
and the binomial theorem to rewrite your congruence as
$$
\tag{$\star$}\frac{1}{p}\sum_{n=0}^{p-1} (1-\zeta^n)^{p^2-p}\equiv p^{p-1}\mod p^p.
$$
The term $n=0$ vanishes, and it is a classical fact that $p|(1-\zeta^n)^{p-1}$ for $1\leq n\leq p-1$. So it is immediate that the left hand side of $(\star)$ is divisible by $p^{p-1}$. With a little work we can compute the residue modulo $p^p$. For $1\leq n\leq p-1$:
$$
\begin{align*}
(1-\zeta^n)^{p-1}&=\sum_{k=0}^{p-1}{p-1\choose k}(-1)^k\zeta^{nk}\\
&=\sum_{k=0}^{p-1}\frac{(1-p)(2-p)\ldots(k-p)}{k!}\zeta^{nk}\\
&=\sum_{k=0}^{p-1}\left(1-\frac{p}{1}\right)\cdots\left(1-\frac{p}{k}\right)\zeta^{nk}\\
&\equiv \sum_{k=0}^{p-1}\left[1-p H_k\right]\zeta^{nk}\\
&\equiv-p\sum_{k=1}^{p-1} H_k \zeta^{nk}\mod p^2,
\end{align*}
$$
where on the last line we used that $\sum_{k=0}^{p-1}\zeta^{nk}=0$.
Here $H_k=1/1 + 1/2 +\ldots+1/k$ is the $k$-th harmonic number. Now we get
$$
\begin{align*}
\frac{1}{p}\sum_{n=1}^{p-1} (1-\zeta^n)^{p^2-p}&=\frac{1}{p}\sum_{n=1}^{p-1}\big((1-\zeta^n)^{p-1}\big)^p\\
&\equiv \frac{1}{p}\sum_{n=1}^{p-1}\left(-p\sum_{k=1}^{p-1} H_k\zeta^{nk}\right)^p\\
&\equiv -p^{p-1}\sum_{n,k=1}^{p-1}H_k^p\zeta^{nkp}\equiv -p^{p-1}\sum_{n,k=1}^{p-1}H_k\\
&\equiv p^{p-1}\sum_{k=1}^{p-1}H_k\equiv p^{p-1}\mod p^p.
\end{align*}
$$
On the third line we used the "Freshman's dream" identity $(\sum x_i)^p\equiv \sum x_i^p\mod p$, along with Fermat's little theorem applied to $H_k$. On the last line we needed $\sum_{k=1}^{p-1}H_k\equiv 1\mod p$, which can be proved by interchanging the order of summation.
