Computing ratio $ \frac{\int_{0}^{1}\left(1-x^{k}\right)^{n} d x}{\int_{0}^{1}\left(1-x^{k}\right)^{n+1}dx} $ by other means than Beta integrals

Question

For any positive numbers $k$ and $n$, converting the integral $$ I_{n}:=\int_{0}^{1}\left(1-x^{k}\right)^{n} d x $$

into a Beta Function by letting $y=x^{k}$, gives $d x=\frac{1}{k} y^{\frac{1}{k}-1} d y$ and $$ \begin{aligned} I_{n} &=\int_{0}^{1} \frac{1}{k} y^{\frac{1}{k}-1}(1-y)^{n} d y \\ &=\frac{1}{k} B \left(\frac{1}{k} , n+1\right) \\ &=\frac{\Gamma\left(\frac{1}{k}\right) \Gamma(n+1)}{k \Gamma\left(\frac{1}{k}+n+1\right)} \end{aligned} $$

Then the ratio becomes $$ \frac{I_{n+1}}{I_{n}}=\frac{\Gamma\left(\frac{1}{k}\right) \Gamma(n+2)}{k \Gamma\left(\frac{1}{k}+n+2\right)} \cdot \frac{k \Gamma\left(\frac{1}{k}+n+1\right)}{\Gamma\left(\frac{1}{k}\right) \Gamma(n+1)} $$

Using the identity $\Gamma(z+1)=z \Gamma(z)$, we can simplify the ratio $$ \frac{\int_{0}^{1}\left(1-x^{k}\right)^{n} d x}{\int_{0}^{1}\left(1-x^{k}\right)^{n+1}dx}=1+\frac{1}{k(n+1)} $$

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By the way, $$ \lim _{n \rightarrow \infty} \frac{\int_{0}^{1}\left(1-x^{k}\right)^{n} d x}{\int_{0}^{1}\left(1-x^{k}\right)^{n+1}dx}= \lim _{k \rightarrow \infty} \frac{\int_{0}^{1}\left(1-x^{k}\right)^{n} d x}{\int_{0}^{1}\left(1-x^{k}\right)^{n+1} d x}=1 $$

My Question: Is there any method other than Beta Function? Your opinion and alternative method is warmly welcome.

Answer 1

Integrate by parts

\begin{align} I_{n+1}=&\int_{0}^{1}\left(1-x^{k}\right)^{n+1}dx\\ =&\>k(n+1) \int_{0}^{1}x^k\left(1-x^{k}\right)^{n}dx = k(n+1) (I_n-I_{n+1}) \end{align} which is $ \frac{I_n}{I_{n+1}}=1+\frac{1}{k(n+1)} $.