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.