= Solution
The <binomial theorem> gives
$$
4^n=(1+1)^{2n}=\sum_{k=0}^{2n}\binom{2n}{k}.
$$
The <central binomial coefficient> is the largest of these $2n+1$ nonnegative <binomial coefficients>, so
$$
\binom{2n}{n}\geq\frac{4^n}{2n+1}=
\frac{2^{2n}}{2n+1}.
$$
For the upper bound, fix a <prime number> $p\leq2n$. By <Legendre formula>,
$$
v_p\!\left(\binom{2n}{n}\right)
=\sum_{j\geq1}
\left(
\left\lfloor\frac{2n}{p^j}\right\rfloor
-2\left\lfloor\frac{n}{p^j}\right\rfloor
\right).
$$
Each summand is either zero or one. Hence the <P-adic valuation> satisfies
$$
v_p\!\left(\binom{2n}{n}\right)
\leq\left\lfloor\frac{\log(2n)}{\log p}\right\rfloor.
$$
If $p\leq\sqrt{2n}$, this implies
$$
p^{v_p(\binom{2n}{n})}\leq2n.
$$
There are at most $\sqrt{2n}$ such <prime numbers>. If instead $p>\sqrt{2n}$, then $p^2>2n$, so the <P-adic valuation> is at most one. Multiplying the prime-power contributions therefore yields
$$
\binom{2n}{n}
\leq(2n)^{\sqrt{2n}}
\prod_{\substack{p\leq2n\\p\text{ prime}}}p.
$$
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