Prove that $\sum_{k=0}^{n}(-1)^k{n\choose k}\frac{1}{(k+1)(k+2)}=\frac{1}{n+2}$ 
Deduce that $\displaystyle {n \choose 0}\dfrac{1}{1\cdot2}-{n \choose 1}\dfrac{1}{2\cdot3}+{n \choose 2}\dfrac{1}{3\cdot4}+...(-1)^n{n \choose n}\dfrac{1}{(n+1)\cdot(n+2)}=\dfrac{1}{n+2}$

We know $\dfrac{1}{n+2}=\displaystyle \int_{0}^{1}t^{n+2-1}dt$
Now $\displaystyle {n \choose 0}\dfrac{1}{1\cdot2}-{n \choose 1}\dfrac{1}{2\cdot3}+{n \choose 2}\dfrac{1}{3\cdot4}+...(-1)^n{n \choose n}\dfrac{1}{(n+1)\cdot(n+2)}=\displaystyle \sum_{k=0}^{n}{n\choose k}(-1)^k\bigg(\dfrac{1}{(k+1)(k+2)}\bigg)=\displaystyle \sum_{k=0}^{n}{n\choose k}(-1)^k\bigg(\dfrac{1}{(k+1)}-\dfrac{1}{k+2}\bigg)$
$=\displaystyle \sum_{k=0}^{n}{n\choose k}(-1)^k\bigg(\dfrac{1}{(k+1)}\bigg)-\displaystyle \sum_{k=0}^{n}{n\choose k}(-1)^k\bigg(\dfrac{1}{(k+2)}\bigg)$
$=\displaystyle \sum_{k=0}^{n}{n\choose k}(-1)^k\int_{0}^1 t^{k+1-1}dt \space - \space \displaystyle \sum_{k=0}^{n}{n\choose k}(-1)^k \int_{0}^1 t^{k+2-1}dt$
$=\displaystyle \int_{0}^1 t^{1-1}\bigg(\sum_{k=0}^{n}{n\choose k}(-t)^k\bigg) dt- \int_{0}^1 t^{2-1}\bigg(\sum_{k=0}^{n}{n\choose k}(-t)^k\bigg) dt$
$=\displaystyle  \int_{0}^1 t^{1-1}(1-t)^{n}dt \space - \space  \int_{0}^1 t^{2-1}(1-t)^{n}dt$
$=\displaystyle \large \beta(1,n+1)$-$\displaystyle \large\beta(2,n+1)$
$=\dfrac{\Gamma(1)\Gamma(n+1)}{\Gamma(1+n+1)}-\dfrac{\Gamma(2)\Gamma(n+1)}{\Gamma(2+n+1)}= \underbrace{\dfrac{1}{n+2}}_{\text{which is exactly what I  proved}}$ 
PS @Chappers Thankyou all users for correcting one nasty mistake.

Hint: 
Multiply and divide by $(n+1)(n+2)$ to turn the question into $$\frac {1}{(n+1)(n+2)} \sum_{k=0}^n (-1)^k \binom {n+2}{k+2}= \frac {1}{(n+1)(n+2)}\left[ \left(\sum_{k=-2}^n (-1)^k \binom {n+2}{k+2}\right)+(n+2)-1\right]=\frac {1}{n+2} $$
