Not only will I prove this, but I will further use the property of subsequences to find the limit of the sequence. 1. We must show that $a_n$ is monotone increasing or decreasing, and 2. that $a_n$ is bounded (above or below). By inspection, $a_1 = 1$, $a_2 = \sqrt{2}$, $a_3 = \sqrt{1+\sqrt{2}}$,... My claim is that the sequence is monotone increasing.
Proof: (by induction on n).
Base case: $n = 1$. Then $a_1 = 1$ and $a_2 = \sqrt{2}$. Since $1<2$, it follows that $\sqrt{1} = 1 < \sqrt{2}$. Thus $a_1 < a_2$
Inductive step: Let $n\in \mathbb{N}$ be given and suppose that $a_{n+1} \geq a_n$. [Goal: $a_{n+2} \geq a_{n+1}$]. Then since $a_{n+1} \geq a_n$, $1+a_{n+1} \geq 1+a_n$ and thus $a_{n+2}=\sqrt{1+a_{n+1}} \geq \sqrt{1+a_n} = a_{n+1}$.
Now we need to show that $a_n$ is bounded above. Proof: (by induction on n)
Base case: $n = 1$, so $a_1 = 1 < 2$.
Inductive Step: Fix $n\in \mathbb{N}$ and suppose $a_n < 2$. [Goal: $a_{n+1} < 2$]. Then, $a_{n+1} = \sqrt{1+a_n} < \sqrt{1+2} = \sqrt{3} < 2$. This is because $4>3$ and thus $2>\sqrt{3}$. Thus $a_{n+1} < 2$ for all $n\in \mathbb{N}$.
Therefore by the MCT, $\{a_n\}_{n\in \mathbb{N}}$ converges.
Well, what does it exactly converge too? Let us say that $a_n$ converges to $L$ where $L\in \mathbb{R}$. We can look at the $a_n$ term to get a better idea. Since, $a_n = \sqrt{1+ a_{n-1}}$, $a_n^2 = 1+a_{n-1}$. Since $a_n$ converges to $L$, it follows that all of its subsequences also converge to $L$ i.e. $a_{n-1}$. Therefore we can rewrite this equation in terms of L. Thus we have that $L^2 = 1 + L$ and that $L^2 - L -1 = 0$. We can use the quadratic formula to obtain the solution that $L = \frac{1+ \sqrt{5}}{2}$. Let me know what you think of this proof! I am not quite sure this L is correct, but let me know about your thoughts. Just trying to get ready for my intro to math proof class final tomorrow.
Your answer is almost perfect. But the quadratic equation gives $L=\frac {1\pm \sqrt 5} 2$. You should observe that $L =\lim a_n \geq 0$ and $\frac {1- \sqrt 5} 2<0$ so we cannot have $L=\frac {1- \sqrt 5} 2$.