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 $k+1$.
Notice OP is the special case $\,n=2k\,$ below, since $\,\color{#c00}{\gcd(k\!+\!1,2k\!+\!1)= 1},\,$ 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}$
$\begin{align} {\bf Proof}\quad q\, :=\,\underbrace{ \dfrac{1}{k\!+\!1}\:\! \dfrac{n!}{k!(n\!-\!k)!}}_{\textstyle \dfrac{1}{\color{#c00}{k\!+\!1}}\Large \ {n \choose k}} &\,=\, \underbrace{\dfrac{1}{n\!+\!1}\:\!\dfrac{(n\!+\!1)!}{(k\!+\!1)!(n\!-\!k)!}}_{\textstyle \dfrac{1}{\color{#c00}{n\!+\!1}}\Large {n+1\choose k+1}}\\[.6em] \end{align}$
Thus $\,q\,$ 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.
Here is a proof using factorials expressing binomial coefficients. Start with the difference of two integers (binomial coefficients) and end up with the Catalan number quotient:
$$ {2n \choose n} - {2n \choose n+1} = \tag{1}$$ $$ \frac{ (2n)! }{n!\, n!} - \frac{(2n)!}{(n+1)!\,(n-1)!} = \tag{2}$$ $$ \frac{ (2n)!\,(n+1)}{n!\,(n+1)\,(n-1)!\,n} - \frac{ (2n)!\,n}{(n+1)!\,(n-1)!\,n} = \tag{3}$$ $$ \frac{ (2n)!}{ (n+1)!\,(n-1)!\,n} \,( (n+1)-n ) = \tag{4}$$ $$ \frac{ (2n)! }{ n!\,(n+1)\,n!}\,(1) = {2n \choose n}\frac1{n+1}. \tag{5}$$
