n-sided dice is rolled n independent times. What is the probability that at least i will be observed on the i-th trial at least once?

My solution:

We will solve it using a complement.

OMEGA = n^n

A' ... A' is an event that any number different from i appears on the i-th trial

A' = (n-1)^i * n^(n-i) ... (n-1)*(n-1)* ...*(n-1)*n*...*n

P(A) = 1 - A'/OMEGA

Official solution:

OMEGA = n^n

A' = (n-1)^n ... A' is an event that any number different from i appears on the i-th trial

P = 1 - A'/OMEGA

I think the official solution is wrong. Because during the first i trials we can observe any number except for i. But in rolls greater than i we can observe any number (even i).

Your thoughts would be greatly appreciated.