Proof Verification: Prove $f(n)<2$, when $f(1)=1$ and $f(n+1) = \sqrt{(2+f(n))}$ for all $n$ positive integers Proceed by induction:
Suppose $P(n)$ is the statement $f(n)<2$, when $n$ $f(1)=1$ and $f(n+1) = \sqrt{(2+f(n))}$ for all $n$ positive integers.
Base case: $P(1)=f(1)=1<2$. True.
Assume $P(n)$, i.e. $f(n)<2$ when $n$ $f(1)=1$ and $f(n+1) = \sqrt{(2+f(n))}$ for all $n$ positive integers, is true.Then $P(n+1)=\sqrt{(2+f(n))}< \sqrt{(2+2)}=2$. QED  
If this is a correct proof, can you please explain how the step:$P(n+1)=\sqrt{(2+f(n))}<\sqrt{(2+2)}=2$ works. Namely, how can we substitute $f(n)$ with $2$.
Thank you! 

Alt. hint:  obviously $f(n) \ge 0$ for all $\,n\,$, then:
$$
\begin{align}
f(n+1) = \sqrt{f(n)+2} \quad&\iff\quad \left(f(n+1)\right)^2 = f(n)+2 \\
 &\iff\quad \left(f(n+1)\right)^2 -4 = f(n) - 2 \\
 &\iff\quad \big(f(n+1)-2\big)\underbrace{\big(f(n+1)+2\big)}_{\gt 0}=f(n)-2
\end{align}
$$
It follows that all differences $f(n)-2$ have the same sign, and since $f(1) \lt 2$ they are all negative.
