Answer
See proof
Work Step by Step
The probability of opening the door with a randomly chosen key is $\frac{1}{n}$ and the probability of not opening the door with a randomly chosen key is $\frac{n-1}{n}$
The first key will not open the door, thus we have $p_1=\frac{n-1}{n}$
The second key will not open the door, thus we have $p_2=\frac{n-2}{n-1}$
The third key will open the door, thus we have $p_3=\frac{1}{n-2}$
The total would be $p=p_1\times p_2\times p_3=\frac{n-1}{n}\times\frac{n-2}{n-1}\times\frac{1}{n-2}=\frac{1}{n}$
In order to get to the third key, the first two keys should not open the door.
