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I thought I'd add another solution that uses recursion. Let $p_n$ be the probability that the n-th person gets the n-th seat (his own seat) and $q_n=1-p_n$ the probability that he does not. Then,

$$ p_n=\underbrace{1/n}_{\text{1-st person sits in 1-st seat}}+\underbrace{(n-2)/n}_{\text{1-st person does not sit the 1-st or n-th seat}} \times p_{n-1} $$ $$ q_n=\underbrace{1/n}_{\text{1-st person sits in n-th seat}}+\underbrace{(n-2)/n}_{\text{1-st person does not sit in the 1-st or n-th seat}} \times q_{n-1} $$ Now substitute in $q_n=1-p_n$ to get $p_n=q_n=1/2$.

Edit: I should add that initially I just thought it was as simple as plugging one equation into the other but indeed an inductive argument is needed like @ely's answer.

I thought I'd add another solution that uses recursion. Let $p_n$ be the probability that the n-th person gets the n-th seat (his own seat) and $q_n=1-p_n$ the probability that he does not. Then,

$$ p_n=\underbrace{1/n}_{\text{1-st person sits in 1-st seat}}+\underbrace{(n-2)/n}_{\text{1-st person does not sit in the 1-st or n-th seat}} \times p_{n-1} $$ $$ q_n=\underbrace{1/n}_{\text{1-st person sits in n-th seat}}+\underbrace{(n-2)/n}_{\text{1-st person does not sit in the 1-st or n-th seat}} \times q_{n-1} $$ Now substitute in $q_n=1-p_n$ to get $p_n=q_n=1/2$.

Edit: I should add that initially I just thought it was as simple as plugging one equation into the other but indeed an inductive argument is needed like @ely's answer.

Edit: my answer assumes that $p_i=p_j$ - basically I haven't taken into account that probabilities might be different depending on which seat the 1-st person sits in (as stated by @hans)... The answer might be salvageable if one instead defines $p_n$ as the probability that everyone on a plane with $n$-seats all sit in their own seat but not sure as this might make some other implicit assumption.

I thought I'd add another solution that uses recursion. Let $p_n$ be the probability that the n-th person gets the n-th seat (his own seat) and $q_n=1-p_n$ the probability that he does not. Then,

$$ p_n=\underbrace{1/n}_{\text{1-st person takes 1-st seat}}+\underbrace{(n-2)/n}_{\text{1-st person takes any seat that is not 1-st or n-th seat}} \times p_{n-1} $$ $$ q_n=\underbrace{1/n}_{\text{1-st person takes n-th seat}}+\underbrace{(n-2)/n}_{\text{1-st person takes any seat that is not 1-st or n-th seat}} \times q_{n-1} $$ Now substitute in $q_n=1-p_n$ to get $p_n=q_n=1/2$.

Edit: I should add that initially I just thought it was as simple as plugging one equation into the other but indeed an inductive argument is needed like @ely's answer.

I thought I'd add another solution that uses recursion. Let $p_n$ be the probability that the n-th person gets the n-th seat (his own seat) and $q_n=1-p_n$ the probability that he does not. Then,

$$ p_n=\underbrace{1/n}_{\text{1-st person sits in 1-st seat}}+\underbrace{(n-2)/n}_{\text{1-st person does not sit the 1-st or n-th seat}} \times p_{n-1} $$ $$ q_n=\underbrace{1/n}_{\text{1-st person sits in n-th seat}}+\underbrace{(n-2)/n}_{\text{1-st person does not sit in the 1-st or n-th seat}} \times q_{n-1} $$ Now substitute in $q_n=1-p_n$ to get $p_n=q_n=1/2$.

Edit: I should add that initially I just thought it was as simple as plugging one equation into the other but indeed an inductive argument is needed like @ely's answer.

I thought I'd add another solution that uses recursion. Let $p_n$ be the probability that the n-th person gets the n-th seat (his own seat) and $q_n=1-p_n$ the probability that he does not. Then,

$$ p_n=\underbrace{1/n}_{\text{1-st person takes 1-st seat}}+\underbrace{(n-2)/n}_{\text{1-st person takes any seat that is not 1-st or n-th seat}} \times p_{n-1} $$

 $$ q_n=\underbrace{1/n}_{\text{1-st person takes n-th seat}}+\underbrace{(n-2)/n}_{\text{1-st person takes any seat that is not 1-st or n-th seat}} \times q_{n-1} $$ Now substitute in $q_n=1-p_n$ to get $p_n=q_n=1/2$.

I thought I'd add another solution that uses recursion. Let $p_n$ be the probability that the n-th person gets the n-th seat (his own seat) and $q_n=1-p_n$ the probability that he does not. Then,

$$ p_n=\underbrace{1/n}_{\text{1-st person takes 1-st seat}}+\underbrace{(n-2)/n}_{\text{1-st person takes any seat that is not 1-st or n-th seat}} \times p_{n-1} $$  $$ q_n=\underbrace{1/n}_{\text{1-st person takes n-th seat}}+\underbrace{(n-2)/n}_{\text{1-st person takes any seat that is not 1-st or n-th seat}} \times q_{n-1} $$ Now substitute in $q_n=1-p_n$ to get $p_n=q_n=1/2$.

Edit: I should add that initially I just thought it was as simple as plugging one equation into the other but indeed an inductive argument is needed like @ely's answer.