Question: $$\frac{1}{1\*2} + \frac{1}{2\*3} + ... + \frac{1}{n(n+1)} = 1 - \frac{1}{n+1}$$

I'm new to induction, but this is what I cam up with so far.

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

Where did I go wrong, or where do I go from here?

EDIT:
After some advice from Patrick (you're awesome!) I have...
$$1 - \frac{1}{(k+1)} + \frac{1}{(k+1)(k+2)} = 1 - \frac{1}{k+2}$$
 $$1 - \frac{k + 2 + 1}{(k+1)(k+2)}$$
 $$1 - \frac{k + 3}{(k+1)(k+2)}$$
 $$\frac{(k+1)(k+2) - (k+3)}{(k+1)(k+2)}$$
 $$\frac{k^2 + 2k + k + 2 - k - 3)}{(k+1)(k+2)}$$
 $$\frac{k^2 + 2k - 1}{(k+1)(k+2)}$$
 $$\frac{k^2 + 2k - 1}{k^2 + 3k + 2}$$

Did I goof up somewhere again?

Thanks.

Answer: 
You went wrong in the first line. You want to prove that $$1-\color{red}{\frac{1}{k+1}} + \frac{1}{(k+1)(k+2)} = 1-\frac{1}{k+2}$$

Indeed, you're trying to show that $$\underbrace{\frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \dots + \frac{1}{k(k+1)}}\_{\text{$1-\frac{1}{k+1}$ by induction hypothesis}} + \frac{1}{(k+1)(k+2)} = 1 - \frac{1}{k+2}$$

Your second line then has a sign error:

$$1 - \frac{1}{(k+1)} + \frac{1}{(k+1)(k+2)} = 1 - \frac{k + 2 \color{red}{-} 1}{(k+1)(k+2)}$$
For every $n\in\mathbb{N}$, let $\mathcal{P}(n)$ the statement
$$
\frac{1}{1\times 2} + \dots + \frac{1}{n(n+1)} = 1 - \frac{1}{n+1}.
$$

1. The statement $\mathcal{P}(1)$ is $\frac{1}{1\times 2} = 1 - \frac{1}{2}$, which is clearly true.
2. We prove that $\mathcal{P}(n) \implies \mathcal{P}(n+1)$ for all $n\geq 1$. We use a direct proof. Suppose
$$
\frac{1}{1\times 2} + \dots + \frac{1}{n(n+1)} = 1 - \frac{1}{n+1}.
$$
Then
$$
\begin{align\*}
\frac{1}{1\times 2} + \dots + \frac{1}{n(n+1)} + \frac{1}{(n+1)(n+2)} 
 &= 1 - \frac{1}{n+1} + \frac{1}{(n+1)(n+2)} \\
 &= 1 - \frac{(n+2) - 1}{(n+1)(n+2)} \\
 &= 1 - \frac{1}{n+2}.
\end{align\*}
$$
This proves that $\mathcal{P}(n) \implies \mathcal{P}(n+1)$.

It follows by induction that
$$
\frac{1}{1\times 2} + \dots + \frac{1}{n(n+1)} = 1 - \frac{1}{n+1}
$$
for all $n\geq 1$.