Proof by induction (how did I do?)

Hi everyone, I'm new to the forums. I'm also fairly new to proofs, and I still commonly make mistakes, specifically regarding quantification errors and circularity. However, I believe that I have gotten this particular proof nailed down pretty well. I was wondering if I could get some feedback on whether or not what I did seems correct. I'm wondering specifically about the fourth line of the induction part, where I said "Note that this equation may also be written as [blah]." Should I elaborate on why, or is it obvious enough?

(Also, sorry about the alignment of some of the longer equations, I can't seem to get it to work like I normally do).

Any help is greatly appreciated!

**Theorem**: For all $\displaystyle n \in \mathbb{N}$, $\displaystyle \frac{1}{1 \cdot 2}+ \frac{1}{2 \cdot 3}+ \cdot \cdot \cdot + \frac{1}{n(n+1)}=\frac{n}{n+1}.$

**Proof**: We will proceed by induction.

i) Base step: Let $\displaystyle n = 1$. We see that $\displaystyle \displaystyle \frac{1}{1 \cdot 2} = \frac{1}{2}$. We also see that

$\displaystyle \frac{n}{n+1} & = \frac{1}{1+1}$

$\displaystyle = \frac{1}{2}.$

Therefore, $\displaystyle P_1$ is true.

ii) Inductive step: Let $\displaystyle k \in \mathbb{N}$. Assume that $\displaystyle \displaystyle \frac{1}{1 \cdot 2}+ \frac{1}{2 \cdot 3}+ \cdot \cdot \cdot + \frac{1}{k(k+1)}=\frac{k}{k+1}$. We would like to show that $\displaystyle \displaystyle \frac{1}{1 \cdot 2}+ \frac{1}{2 \cdot 3}+ \cdot \cdot \cdot + \frac{1}{(k+1)(k+2)}=\frac{k+1}{k+2}$. Note that this equation may also be written as $\displaystyle \displaystyle \frac{1}{1 \cdot 2}+ \frac{1}{2 \cdot 3}+ \cdot \cdot \cdot + \frac{1}{k(k+1)} + \frac{1}{(k+1)(k+2)}=\frac{k+1}{k+2}$. By our assumption, we see that

$\displaystyle \displaystyle \frac{1}{1 \cdot 2}+ \frac{1}{2 \cdot 3}+ \cdot \cdot \cdot + \frac{1}{k(k+1)} + \frac{1}{(k+1)(k+2)} & = \frac{k}{k+1} + \frac{1}{(k+1)(k+2)}$

$\displaystyle & = \frac{k}{k+1}\Bigg(\frac{k+2}{k+2}\Bigg) + \frac{1}{(k+1)(k+2)}$

$\displaystyle = \frac{k^2+2k+1}{(k+1)(k+2)}$

$\displaystyle = \frac{(k+1)^2}{(k+1)(k+2)}$

$\displaystyle = \frac{k+1}{k+2}.$

Therefore, $\displaystyle P_{k+1}$ is true.

Furthermore, by the PMI, $\displaystyle P_n$ is true for all $\displaystyle n \in \mathbb{N}$.