What is the probability of getting a heart in $n$-th draw without replacement?

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You are given a deck of cards. What is the probability of getting a heart at the first draw. What is the probability of getting the same in second draw without replacement? What is the probability of getting the same in $n$-th draw without replacement? Give explanation of your answer.

The probability of getting a heart in the first draw =$\dfrac{13}{52}=\dfrac14$

Using theorem of total probability the probability of getting a heart in the second draw =$\dfrac{13}{52}\times\dfrac{12}{51}+\dfrac{39}{52}\times \dfrac{13}{51}=\dfrac14$

So, intuitively the probability of getting a heart in $n-$th draw will be $\dfrac14$. But I am unable to give the explanation. Please help.

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Suppose that we put cards on the bottom of the deck after taking them from the top. Then the $n^\text{th}$ card drawn is the top card, which we know has probability 1/4 of being a heart. (The point is that which card is considered to be the top one is arbitrary and meaningless, so it has no affect on the probability.)