Proof of divisibility with combinatorics

Prove that, for every positive integer $\,k, {2k\choose k}\,$ is divisible by $k+1$. Hint: $\frac{2k+1}{k+1}{2k\choose k}$

What I have done so far

${2k\choose k} = 2k!/k!k!=((k+k)(k+k+1)(k+k+2)\cdots k!)/k!k!= ((k+1)(2k!/(k+1))/k!k!$

Is this proof correct or is there a better way to prove it?

You are correct that $\binom{2k}{k} = \frac{(k+1) (2k)! / (k+1)}{k! k!}$, but this doesn't show that $\binom{2k}{k}$ is divisible by $k+1$. You can take any number $n$ and write $n = n(k+1)/(k+1)$ even though $n$ is not divisible by $k+1$.

The purpose of the hint is for you to recognize that $\frac{2k+1}{k+1} \binom{2k}{k} = \binom{2k+1}{k}$ is an integer. This means $(2k+1) \binom{2k}{k}$ is divisible by $k+1$. To finish the proof, think about the common factors of $2k+1$ and $k+1$.

It is the special case $\,n=2k\,$ below, by $\,\color{#c00}{\gcd(k\!+\!1,2k\!+\!1)= 1},\,$ by Euclid.

Lemma $\ \ $ Both $\ \ k\!+\!1\mid {n\choose k}\ \ $ and $\ \ n\!+\!1\mid {n+1\choose k+1}\ $ when $ \ \color{#c00}{\gcd(k\!+\!1,n\!+\!1)=1}$

$\begin{align} {\bf Proof}\qquad \dfrac{1}{k\!+\!1}\:\! \dfrac{n!}{k!(n\!-\!k)!} &\,=\, \dfrac{1}{n\!+\!1}\:\!\dfrac{(n\!+\!1)!}{(k\!+\!1)!(n\!-\!k)!}\\[.6em] \Rightarrow\ \ \ q\, :=\, \dfrac{1}{\color{#c00}{k\!+\!1}}\ \ \ {n \choose k}\ \ \ =\!\!\! &\qquad\!\dfrac{1}{\color{#c00}{n\!+\!1}}\,{n\!+\!1\choose k\!+\!1} \end{align}$

Therefore $\,q\,$ is an integer, being a fraction writable with $\rm\color{#c00}{coprime}$ denominators.

See here for the straightforward generalization to the non-$\rm\color{#c00}{coprime}$ case.

Remark $ $ Conceptually, the order of $\,q\,$ (= least denominator) divides coprimes so it must be $1,\,$ where the order is interpreted in the group $\,\Bbb Q/\Bbb Z = $ rationals $\!\bmod 1.\,$ Equivalently, said more simply: it is well known since grade school that the least denominator of a fraction divides every other denominator (though this is usually not proved till much later, e.g. by Euclid's Lemma or unique prime factorization); thus since the least denom divides coprimes it must be $1$. The prior link gives proof(s) of this property of fractions, and elaborates on the various conceptual viewpoints emphasized above.