Let us count the number of possible arrangements under this scheme with $n>1$ people, because for $n=1$, the probability that the $n^{th}=1^{st}$ person gets their own seat is $1$, that being the only possibility.
For $n>1$, we want to count the number of scenarios when the $n^{th}$ person will be sitting in his own seat and divide that by the total number of possible arrangements. So, we have seats $S_1,S_2,\cdots , S_n$ reserved for passengers $P_1,P_2,\cdots , P_n$. In our favourable arrangement $P_n$ is occupied by $S_n$.
So, $P_1,\cdots, P_{n-1}$ must have found seats in $S_1,\cdots,S_{n-1}$ only.
Suppose after entering the plane, $P_1$ chose $S_5$.
Then $P_2,P_3,P_4$ will have their seats empty and sit in $S_2,S_3,S_4$ respectively. $(*)$
But $P_5$ sees $S_5$ occupied, so he could have chosen $S_1$ to sit on, or one of the seats from $S_6,\cdots,S_{n-1}$.
Case $1$: If $P_5$ chooses $S_1$, then everyone from $P_6$ to $P_{n-1}$ will find their respective seats empty and sit there, and the only disturbance in seating arrangement (i.e. not sitting in respective reserved places) would be caused by the people with indices $(1,5)$.
Case $2$: If $P_5$ chooses a seat other than $S_1$, say $S_{17}$, then $P_6,\cdots,P_{16}$ follow the same procedure in $(*)$ and $P_{17}$ has to make the decision similar to Case 1: or Case 2:, and again if $P_{17}$ chose $S_1$ to sit in, then $P_{18},\cdots,P_{n-1}$ sit happily in $S_{18},\cdots,S_{n-1}$, so that the disturbance mentioned in the above block is caused by people with indices in $(1,5,17)$ or we come to another Case 2: with $P_{17}$ in place of $P_5$.
Thus, we see that specifying a subset $A\subseteq \{2,3,4,\cdots,n-1\}$, for example $A=\{5,17,33\}$, the people in the set $A\cup \{1\}=\{1,5,17,33\}$ correspond to the seating arrangement, where $$P_1\rightarrow S_5, P_5\rightarrow S_{17}, P_{17}\rightarrow S_{33}, P_{33}\rightarrow S_{1}$$ and the rest are seating in their own places due to the logic in $(*)$. The empty subset $A=\phi$ corresponding with $A\cup \{1\}=\{1\}$ indicates the seating arrangement that $$P_1\rightarrow S_1$$ so that everyone else is seated in their reserved places.
Thus we have a bijection, each of the favourable arrangements corresponds to a unique subset of $\{2,3,\cdots,n-1\}$, so there are as many favourable arrangements as there are subsets of $\{2,3,\cdots,n-1\}$ (of size $n-2$), which is $2^{n-2}$.
The number of total possible outcomes, is the exact same analysis as above, with the possible seats being $S_1,\cdots,S_n$ for $P_1,\cdots,P_n$ so that we have to consider all the subsets of $\{2,3,\cdots,n\}$, which are $2^{n-1}$ in number.
So the probability for $n>1$ passengers is $\dfrac{2^{n-2}}{2^{n-1}}=\dfrac12$.
