show that $(n+1)(n+2)...(2n)$ is divisible by $2^n$ but not by $2^{n+1}$

Note that $$\dfrac{(n+1)(n+2)\dots (2n)}{1\cdot 3 \cdot 5 \dots (2n-1)} = \dfrac{(2n)!/n!}{(2n)!/(2\cdot 4 \cdot 6 \cdot \dots \cdot (2n))} = \dfrac{(2n)!/n!}{(2n)!/(2^n n!)} = 2^n$$

Hence, $$(n+1)(n+2)\dots (2n) = 1\cdot 3 \cdot 5 \dots (2n-1) \cdot 2^n$$

Since the first $n$ multipliers on the right side are all odd, the maximum power of two that divides the product in question is indeed $2^n$.

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Here is some intuition behind this solution.

First of all, it is clear that every natural number $m$ can be represented in the form $$m = 2^k \operatorname{maxodd}(m)$$ where $2^k$ is the maximum power of two that divides $m$ and $\operatorname{maxodd}(m)$ is the maximum odd divisor of $m$.

Note that the numbers in the range $[n+1, 2n]$ have different maximum odd divisors. Indeed, if for two different numbers in this range their maximum odd divisors were the same, one of them should be at least two times greater than the other one, which is impossible. This means that each of the number $1, 3, 5, \dots, 2n-1$ (here are exactly $n$ numbers) should be the maximum odd divisor for exactly one number from $[n+1, 2n]$ range.

For example, if $n=5$ then $[n+1, 2n] = \{6, 7, 8, 9, 10\}$ and maximum odd divisors are $\{3, 7, 1, 9, 5\} = \{1, 3, 5, 7, 9\}$.

This means that the power of two in the product is equal to

$$\dfrac{(n+1)(n+2)\dots (2n)}{1\cdot 3 \cdot 5 \dots (2n-1)} = \dfrac{(2n)!/n!}{(2n)!/(2\cdot 4 \cdot 6 \cdot \dots \cdot (2n))} = \dfrac{(2n)!/n!}{(2n)!/(2^n n!)} = 2^n$$

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \pars{n + 1}\cdots\pars{2n} & = {\pars{2n}! \over n!} = {\pars{2n}\pars{2n - 1}\pars{2n - 2}\pars{2n - 3}\cdots 3.2 \over n!} = 2^{n}\prod_{k = 1}^{n}\pars{2k - 1} \end{align}

You can also proceed by induction.

Let $a_n = (n+1)(n+2)\ldots(2n)$.

Then $a_{n+1}/a_n = (2n+1)(2n+2)/(n+1) = 2(2n+1)$.
Since $2n+1$ is odd, only one $2$ factor is added at each step.

Since $a_1 = 2^1$, you get $a_n = 2^n \times o_n$ where $o_n$ is an odd number