Question: I have some difficulties proving the following statement: $\sqrt {n+1} - \sqrt n ≤ 1$

Thus far I have the following:

The root of a positive integer is always smaller than or equal to the integer, since $a^2 ≥ a$ where a is a positive integer.

Therefore $\sqrt {n+1} ≤ \sqrt n + 1$ and thus $\sqrt {n+1} - \sqrt n ≤ 1$

However I feel like I am jumping to a conclusion and I'd like to be more precise.

This form would be precise if I'd want to prove any of the following:

$\sqrt {n+1} ≤ n + 1$

or
$\sqrt n + \sqrt 1 ≤ \sqrt n + 1$

But I don't think it is precise enough for what I actually want to prove. Any ideas?
I understand the statement is true. Adding an integer $x$ to an integer $n$ and then taking the square root of that will always be less than the squre root of $n$ + $x$, I just don't know how to prove it.

Answer: 
write $$\sqrt{n+1}\le 1+\sqrt{n}$$ after squaring (all summands are positive) we get
$$0\le \sqrt{n}$$ which is true.
Not exactly number-theoretic, but for $n=0$ the inequality is obvious, while for $n\ge1$ we have

$$\sqrt{n+1}-\sqrt n=\int\_n^{n+1}{2dx\over\sqrt x}\le\int\_n^{n+1}{dx\over2\sqrt n}={1\over2\sqrt n}\le1$$