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.
There are several ways to prove this. Here is an algebraic one. Start with the difference of two integers (binomial coefficients) and end up with the Catalan number quotient:
$$ {2n \choose n} - {2n \choose n+1} = \tag{1}$$ $$ \frac{ (2n)! }{n!\, n!} - \frac{(2n)!}{(n+1)!\,(n-1)!} = \tag{2}$$ $$ \frac{ (2n)!\,(n+1)}{n!\,(n+1)\,(n-1)!\,n} - \frac{ (2n)!\,n}{(n+1)!\,(n-1)!\,n} = \tag{3}$$ $$ \frac{ (2n)!}{ (n+1)!\,(n-1)!\,n} \,( (n+1)-n ) = \tag{4}$$ $$ \frac{ (2n)! }{ n!\,(n+1)\,n!}\,(1) = {2n \choose n}\frac1{n+1}. \tag{5}$$
