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.
0) For $n=0:$ $(1)^{1/2}-0 =1$.

1) Consider $n \ne 0$.

$f(x) = x^{1/2}$ is continuos in $[n,n+1]$, differentiable in $(n,n+1)$.

Mean value Theorem:

$\dfrac{f(n+1) -f(n)}{1} = f'(t)$, 

where $t \in (n,n+1)$.

$(n+1)^{1/2} - (n)^{1/2} =$

$ (1/2)(t)^{-1/2} \lt 1$,

where $t \in (n,n+1)$.