Limit of a recursively defined sequence

I'm having some trouble with this question.

Suppose that $\displaystyle a_1=1$ and $\displaystyle a_{n+1}=\sqrt{1+a_n}$ whenever $\displaystyle n\geq 1$

1) Show that $\displaystyle \sqrt{1+x}\in [1,2]$ whenever $\displaystyle x\in [1,2]$.

2) Use induction to show that the sequence $\displaystyle \{a_n\}$ is bounded.

Would I do the question like this:

Prove that $\displaystyle a_n<2 \ \ \ \forall n\geq 1$

$\displaystyle a_1=1<2$

Assume $\displaystyle a_k<2$, then $\displaystyle a_{k+1}=\sqrt{1+a_k}<\sqrt{3}<2$

By induction it follows that $\displaystyle a_n<2 \ \ \ \forall n\geq 1 \Rightarrow a_n$ is bounded above by 2.

3) Use induction to show that $\displaystyle {a_n}$ is an increasing sequence

Prove $\displaystyle a_{n+1}>a_n \ \ \ \forall n\geq 1$

$\displaystyle a_2=\sqrt{1+1}=\sqrt{2}>a_1=1$

Assume that $\displaystyle a_{k+1}>a_k$

$\displaystyle a_{k+2}=\sqrt{1+a_{k+1}}>\sqrt{1+a_k}=a_{k+1}$

By induction it follows that, $\displaystyle a_{n+1}>a_n \ \ \ \forall n\geq 1 \Rightarrow a_n$ is an increasing sequence

4) Explain why $\displaystyle \lim_{x \to \infty}a_n$ exists

Since $\displaystyle a_n$ is a monotonically increasing sequence, bounded above, it must converge to a limit as $\displaystyle n\rightarrow\infty$

5) Find $\displaystyle \lim_{x \to \infty}a_n$. (That is, find $\displaystyle \sqrt{1+\sqrt{1+\sqrt{1+...}}}$

Can someone check whether what I did is correct and help me with (1) and (5)