I am not sure if this is good, so I would like someone to help me a little and tell me if this is a good proof.
I know how to prove that, for example, $\displaystyle \sqrt{2}$ is irrational, so I tried to do something similar with this expression.

So, let's assume otherwise, that $\displaystyle \sqrt{n+\sqrt{n}}$ is not irrational, then it must be rational ie. $\displaystyle \sqrt{n+\sqrt{n}}\in\mathbb{Q}$. Then we can write $\displaystyle \sqrt{n+\sqrt{n}}=\dfrac{a}{b}, \, a\in\mathbb{Z}, b\in\mathbb{N}$. After squaring the expression, we have: $\displaystyle \sqrt{n}=\dfrac{a^2}{b^2}-n$. Since $\displaystyle \mathbb{N}\subset\mathbb{Q}$, then also $\displaystyle n\in\mathbb{Q}$. And since there is closure of subtraction in $\displaystyle \mathbb{Q}$, then $\displaystyle \dfrac{a^2}{b^2}-n\in\mathbb{Q}$. So it is obvious that it must be $\displaystyle \sqrt{n}\in\mathbb{Q}$. If n is not a perfect square then it is obvious that
$\displaystyle \sqrt{n}\notin\mathbb{Q}$, so we have a contradiction. Then, let's assume that n is a perfect square then $\displaystyle \sqrt{n}\in\mathbb{Q}$, and we can write $\displaystyle n=k^2, \, k\in\mathbb{Q}$. Let's also assume that k is not a perfect square, then we have: $\displaystyle \dfrac{a}{b}=\pm\sqrt{k(k+1)}$. But we assumed that k is not a perfect square so $\displaystyle \sqrt{k}$ is irrational and therefore $\displaystyle \dfrac{a}{b}$ is irrational which is in contradiction with assumption that $\displaystyle \dfrac{a}{b}$ is rational. But also, if k is a perfect square, then it is obvious that $\displaystyle k+1$ is not a perfect square, so $\displaystyle \sqrt{k+1}$ is irrational. Then again, $\displaystyle \dfrac{a}{b}$ is irrational which is in contradiction with assumption that $\displaystyle \dfrac{a}{b}$ is rational. So, $\displaystyle \sqrt{n+\sqrt{n}}$ can't be rational, it must be irrational.