If $n = mp^e$ where $e$ is maximal, then $\binom{n}{p^e}$ is not divisible by $p$. Let $n \geq 2$ be an integer, $p$ a prime with $p^e$ the highest power of $p$ dividing $n$.  Then $\binom{n}{p^e}$ is not divisible by $p$.
I think you can do it using this formula for $\binom{n}{k}$.  From that post we have that
$$
\binom{n}{r} = (\prod_{n-r\lt p\leq n} p^{k_p(n)})(\prod_{r\lt p\leq n-r} p^{k_p(n) - k_p(n-r)})(\prod_{p\leq r}p^{k_p(n)-k_p(n-r)-k_p(r)})
$$
Well $p$ falls into the class $p \leq r= p^e$.  So it suffices to show that $k_p(n) - k_p(n-p^e) - k_p(p^e) = 0$ where $k_p(n) = \sum_{k\geq 1} \lfloor \frac{n}{p^k}\rfloor$.  Well, $k_p(n) = k_p(mp^e) = m + mp + mp^2 + ... mp^{e-1}$, and 
$$
k_p(n - p^e) = k_p((m - 1)p^e)
$$
but $m-1$ could equal $\ell p^k$ for some $k$.  So I'm not sure where to go from here.  Alternative simpler proofs welcome.
The proof from the book goes like this:
For each $n-k$ divisible by $p$ in the numerator of $\binom{n}{p^e}$, $n-k = mp^e - \ell p^i$.  It says "then $i \lt e$", but I don't how that's true.  Without that requirement we have either $e = i, \ e \gt i,$ or $e \lt i$.  Let's see.  If $e \leq i$, then $p^e - k = p^e(1 - \ell p^{i-e})$, but there is not such $p^e-k$ in the denominator so $e \gt i$. It then says $(m-k) = (p^e - k)$ and $(n-k) = (p^em - k)$ are both divisible by $p^i$ but not $p^{i+1}$.  How did they get $m-k = p^e - k$?
 A: This is a direct consequence of Lucas' Theorem.
Write $n$ and $p^e$ as base-$p$ numbers, $n$ would have exactly $e$ $0$'s at the end, and would have a digit $1\le d< p$ to the left of those $0$'s. $p^e$ also has exactly $e$ zeros,  with a leading $1$. 
Clearly, the binomial coefficients in the Lucas' theorem are all ones, except for the $(e+1)$-th from the right, which is $\binom{d}{1}=d<p$, so $\binom{n}{p^e}\equiv d\not\equiv 0 \pmod{p}$.
A: Personally I would use Lucas' theorem for binomial coefficient residues in $p$-digital expansions:
$$\binom{a_kp^k+\cdots+a_1p+a_0}{b_kp^k+\cdots+b_1p+b_0}\equiv\binom{a_k}{b_k}\binom{a_{k-1}}{b_{k-1}}\cdots\binom{a_1}{b_1}\binom{a_0}{b_0}\mod p.$$
(Here $0\le a_i,b_j<p$.) The theorem telescopes out inductively from the following version:
$$\binom{ap+r}{bp+s}\equiv\binom{a}{b}\binom{r}{s}\mod p,$$
where $0\le r,s<p$ and $a,b\ge0$. This follows from using the binomial theorem and freshman's dream in characteristic $p$. In "generatingfunctionological" language, we can observe
$$\sum_{b,s} \binom{ap+r}{bp+s}X^{bp+s}\equiv(1+X)^{ap+r}\equiv(1+X^p)^a(1+X)^r\equiv\sum_{b,s} \binom{a}{b}\binom{r}{s}X^{ap+r},$$
with $b$ ranging as $0\le b\le a$ and $s$ ranging as $0\le s<p$ if $b<a$ and $0\le s\le r$ if $b=a$.
Equating coefficients gives Lucas' theorem. Afterwards we have
$$\binom{mp^e}{p^e}\equiv\binom{m}{1}\equiv m\not\equiv0\mod p.$$
