Solution
= Solution
Because $p\geq k+2$, neither $(k+1)!$ nor $k!$ contains a factor $p$. On the other hand, $p\leq2k+1$, so $(2k+1)!$ contains a factor $p$; and
$$
2p\geq2k+4>2k+1,
$$
so it contains no second factor $p$. Therefore
$$
v_p\left(\binom{2k+1}{k+1}\right)
=v_p((2k+1)!)-v_p((k+1)!)-v_p(k!)
=1.
$$
This is the <valuation of a near-central binomial coefficient>.