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 $k+1$.
Conceptually the hint solves it immediately (no calculation!) by combining two basic facts:
$(1)\,$ A fraction $\:\!q\:\!$ is an integer if it is writable two ways with $\rm\color{#c00}{co}\color{#0a0}{prime}$ denominators, by here.
$(2)\,$ $\displaystyle\small \frac{\color{#0a0}{n\!+\!1}}{k\!+\!1}{n\choose k}\overset{\rm\color{darkorange}{B_{\vphantom{|}}}} = {n\!+\!1\choose k\!+\!1}\ $ so $\, q={\small\displaystyle\frac{1}{\color{#c00}{k+1}}{2k\choose k}}$ has denoms $\:\!\color{#c00}{k\!+\!1},\:\!$ $\:\!\color{#0a0}{2k\!+\!1},\,$ by $\,\underbrace{(\color{#0a0}{2k\!+\!1})\:\!q\in\Bbb Z}_{\textstyle\small {\rm by}\,\ \color{#0a0}{n\!=\!2k}\,\ {\rm in}\,\ \rm\color{darkorange}{B^{\vphantom{|}}}} $ Remark $ $ As explained in said answer, conceptually, the order of $\:\!q\:\!$ (= least denominator) divides coprimes so it must be $1\,$ (working in the additive rational circle 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, so since the least denom divides coprimes it must be $1$.
Said answer gives various proofs of this fundamental property fractions and its natural extension - that possible denominators for a fraction are closed under gcd - and elaborates further on various conceptual viewpoints. This basic property proves quite handy but often goes without mention in most elementary number theory textbooks, so it deserves further emphasis.
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:
$$ {2n \choose n} - {2n \choose n+1} = \tag{1}$$ $$ \frac{ (2n)! }{n!\, n!} - \frac{(2n)!}{(n+1)!\,(n-1)!} = \tag{2}$$ $$ \frac{ (2n)!\,(n+1)}{n!\,(n+1)\,(n-1)!\,n} - \frac{ (2n)!\,n}{(n+1)!\,(n-1)!\,n} = \tag{3}$$ $$ \frac{ (2n)!}{ (n+1)!\,(n-1)!\,n} \,( (n+1)-n ) = \tag{4}$$ $$ \frac{ (2n)! }{ n!\,(n+1)\,n!}\,(1) = {2n \choose n}\frac1{n+1}. \tag{5}$$
Let me add a combinatorial proof to the other excellent ones. I am aware that you were required to prove the statement in a different way.
Let $A$ be a set with $2 k$ elements. Given any of the $\dbinom{2 k}{k-1}$ subsets of $A$ with $k-1$ elements, there are $k+1$ ways to complete it to a subset of $A$ with $k$ elements. This way, each of the $\dbinom{2 k}{k}$ subsets of $A$ with $k$ elements has been counted $k$ times, once for each element you may drop from it to get a subset of $A$ with $k-1$ elements. This yields $$\tag{identity} \binom{2 k}{k} \cdot k = \dbinom{2 k}{k-1} \cdot (k + 1). $$ Thus $k+1$ divides LHS of (identity}. Since $k + 1$ is coprime to $k$, it follows that $k + 1$ divides $\dbinom{2 k}{k}$.