Dividing by $p^{n-1}$ we need to prove that: ${p^{n}-1\choose p-1}\equiv 1 \ (mod p)$ By Wilson we get $b=(p-1)!\equiv -1 \ (mod p)$ Also $a=\frac{(p^{n}-1)!}{(p^{n}-p)!}\equiv (p-1)!\equiv -1 \ (modp)$ Now we need to prove that $\frac{a}{b}\equiv 1 \ (modp)$ Obviously $\exists_{k\in \mathbb{N}} \ a=k\cdot b$ and $(b,p)=(kb,p)=1$ But $p|a-b=kb-b=(k-1)b$ so $k\equiv 1 \ (modp)$ proving the thesis Regards Piotr Rutkowski

$\prod_{k=1}^{p-1} (p^n - k) - (p-1)! \equiv (-1)^{p-1} (p-1)! - (p-1)! = 0 \pmod{p}$. No need for Wilson's theorem.

Exactly $p^{2n+v(p)}|\binom{p^n}{p}-p^{n-1}$, were $v(2)=-1, v(3)=0, v(p)=1, p\not |B_{p-3}, p>3$ and $v(p)=2,p|B_{p-3}.$

See here also http://www.artofproblemsolving.com/Forum/viewtopic.php?f=56&t=227653

Using the identity ${n\choose k}=\frac n k {{n-1}\choose {k-1}}$ we find that it suffices to prove $p|{{p^n-1}\choose {p-1}}-1$.Now we get $(p^n-1).......(p^n-p+1)-(p-1)!\equiv (p-1)!-(p-1)!\equiv 0\mod p$ and we are done since $gcd((p-1)!,p)=1$

WakeUp wrote: Prove that for each prime number $p$ and positive integer $n$, $p^n$ divides \[\binom{p^n}{p}-p^{n-1}. \] $${p^n \choose p} = \frac{p^n (p^n-1)(p^n-2) \cdots (p^n-p+1)}{p!} - p^{n-1} = p^{n-1} \left \{ \cdot \frac{(p^n-1)(p^n-2) \cdots (p^n-p+1)}{(p-1)!} - 1 \right \}$$so it suffices to show that $$ p \mid \frac{(p^n-1)(p^n-2) \cdots (p^n-p+1)}{(p-1)!} - 1$$or $$p \mid (p^n-1)(p^n-2) \cdots (p^n-p+1) - (p-1)!$$which is obvious.

A combinatorial way is this: From the set of first $p^n$ naturals , we can select $p$ of them in ${p^n \choose p}$ ways. But now, we can group such $p$-element subsets in equivalence classes in the following way: $\{a_1 , a_2 , a_3 , \cdots , a_p\} \equiv \{b_1 , b_2 , b_3 , \cdots , b_p\}$ if there exists a $0 \leq t < p^n$ such that $$\{(a_1+t) \pmod {p^n}, (a_2+t) \pmod {p^n} , \cdots , (a_p+t) \pmod {p^n} \} = \{b_1 , b_2 , b_3 , \cdots , b_p\}$$ For example if $p = 3$ and $ n = 3$, then the $3$-element subset $\{1,5,26\}$ is in the same equivalence class as $\{2,6,27\}$ , $\{3,7,1\}$, $\cdots$ , $\{27,4,25\}$. It is not hard to see that most equivalence classes have $p^n$ elements; the only equivalence classes that do not have $p^n$ elements are those whose elements are of the form $$\{a , a+p^{n-1} , a+2p^{n-1}, \cdots , a+ (p-1)p^{n-1}\}$$There are $p^{n-1}$ such subsets. Thus $$p^n \mid {p^n \choose p} - p^{n-1}$$

Notice $\binom{p^n}{p} = \frac{(p^n)!}{p!\cdot (p^n-p)!} = p^{n-1} \cdot \binom{p^n-1}{p-1}$. We have \[\binom{p^n-1}{p-1} - 1\equiv (-1)^{p-1}\cdot \frac{(p-1)!}{(p-1)!} -1 \equiv 1 - 1\equiv 0 \pmod p\]and we are done.