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Maybe it's a good idea to start from a situation that's intuitive, and then gradually chance it to the Monty Hall version.

So let's start with the following situation:

Monty lets you choose a door. Then, after you've chosen it, he offers you to either select the chosen door, or open *all* the other doors. Well, it should be clear to everyone that staying at the one door gives you only a chance of $1/n$ to get the car, while opening all the other doors gives you a chance of getting the car. Note that if the car is behind those $n-1$ doors, opening all those doors gives you an additional bit of information, but one that is irrelevant to you: Behind *which* of those doors the car sits.

Now let's assume that in order to save time, Monty does not open all those doors, but only the one with the car if it is among them, and a random door with a goat otherwise. Obviously that doesn't affect the probability of winning the car.

Now let's, on the contrary, assume Monty wants to increase the tension by first opening all those of the doors neither you nor he did choose. In other words, he first reveals $n-2$ goats. Since you know in advance that those $n-2$ doors will show goats, this gives you no information about whether you've won the car. It *does* give you some information about in which of those doors the car is in case it is behind one of them, but as noted above, that information is irrelevant. Therefore your probability of having won the car still stays the same, as you didn't get any new information about whether you've won it.

But if opening those doors doesn't give you any relevant information, we can move it to before you made the choice without affecting the probabilities. That is, Monty will first open the $n-2$ goat doors, and then present you with the choice whether you want to stay with the door you chose, or select the other $n-1$ doors. But since $n-2$ doors are already open, choosing the $n-1$ doors is equivalent to you choosing the one of them which hasn't been opened.

But now we've arrived at the original Monty Hall situation. And since at no point during the transformation the probabilities changed, you still have the probability of $(n-1)/n$ if switching versus $1/n$ if staying.