What whole number $n$ satisfies: $ n<\sqrt{42+\sqrt{42+\sqrt{42}}} < n+1 $ Problem

As the title says: What whole number $n$ satisfies $$ n<\sqrt{42+\sqrt{42+\sqrt{42}}} < n+1 $$

My attempt at answer, which is not rigorous at all
Let's try to calculate the middle part. We make the approximation that $\sqrt{42} ≈ 7$. Substituting this value gives:
$\sqrt{42+\sqrt{42+7}} = \sqrt{42+\sqrt{49}}=\sqrt{42+7}=\sqrt{49}=7 $
However, since we approximated this doesn't really equate to $7$. But the actual value must be between $6$ and $7$. Let's call the actual number $a, \: a\in]6;7[$
Now the inequality looks like this:
$$ n < a < n+1, \: \: a \in]6;7[$$
From here it's easy to see that $n=6$ is the only whole number that satisfies the inequality.
My question
My attempt at an answer wasn't very analytic or rigouros, nor do I know if it is right. Is there a way to solve the problem more elegantly that doesn't rely on my silly idea of bad approximations?
 A: Your solution is correct:

*

*Since $\sqrt{42}<7$, $\sqrt{42+\sqrt{42+\sqrt{42}}}<7$.


*Since $\sqrt{42}>6$, $\sqrt{42+\sqrt{42}}>\sqrt{48}>6$,
so $\sqrt{42+\sqrt{42+\sqrt{42}}}>\sqrt{48}>6$.
So $n=6$.
A: Hint:  $6 < \sqrt{42} < 7$. Use this to find $n$.
A: Set $x$ equal to the infinite nested root $\sqrt{n+\sqrt{n+\sqrt{n+...}}}$
Then $x^2=n+x$, so we have $x\cdot(x-1)=n$
So, whenever $n$ is the product of two consecutive integers, the infinite radical is equal to the larger of the two integers.
It so happens that $42=6\times7$, therefore $\sqrt{42+\sqrt{42+\sqrt{42+...}}}=7$
Since your expression is not infinite, but is truncated, that means the value of your expression is less than $7$, and it is clearly greater than $6$ because $\sqrt{42}>6$.
A: Can we check this: I think that $6<\sqrt{42+\sqrt{42+\sqrt{42}}}$ is obvious, but
$\sqrt{42+\sqrt{42+\sqrt{42}}}<7$ isn't. But
\begin{align}
\sqrt{42+\sqrt{42+\sqrt{42}}}<7&\iff 42+\sqrt{42+\sqrt{42}}<49\\
&\iff \sqrt{42+\sqrt{42}}<7\\
&\iff 42+\sqrt{42}<49\\
&\iff \sqrt{42}<7\\
&\iff 42<49
\end{align}
which is true. And we can do this for any number of nestings with $\sqrt{42}$.
