I am solving a problem which asks for deriving the formula $$\lim_ {n\to\infty} \left(\frac{((2r+1)n+1)!} {{n!}^{2r+1} } \right)^{1/n} = (2r+1)^{2r+1}.$$
I tried using Stirling Approximation which is $$n!\sim(2πn)^{1/2}(n/e)^n \text{ as }n\to\infty.\tag1$$
My attempt: I first took $n=((2r+1)n+1)$ in (1) and then $n=n$ in (1) and then using the fact that $n\to\infty$ . But finally I am getting $\frac{((2r+1)n+1)^{2r+1}}{n ×e^{2r+1}}$ , instead of what should be the answer.
I know this question is a bit of calculation but still if someone can show exact calculations I shall be really thankful.
Using the root-to-ratio limit
$$\lim_{n\to\infty}\sqrt[n]{a_n}=\lim_{n\to\infty}\frac{a_{n+1}}{a_n}$$
we get
\begin{align}L&=\lim_{n\to\infty}\frac{((2r+1)(n+1)+1)!}{(n+1)!^{2r+1}}\frac{n!^{2r+1}}{((2r+1)n+1)!}\\&=\lim_{n\to\infty}\frac{n!^{2r+1}}{(n+1)!^{2r+1}}\frac{((2r+1)n+2r+2)!}{((2r+1)n+1)!}\\&=\lim_{n\to\infty}\frac1{(n+1)^{2r+1}}\prod_{i=2}^{2r+2}((2r+1)n+i)\\&=\lim_{n\to\infty}\prod_{i=2}^{2r+2}\frac{(2r+1)n+i}{n+1}\\&=\prod_{i=2}^{2r+2}(2r+1)\\&=(2r+1)^{2r+1}\end{align}
as claimed.