Why is the norm of $x_n(t) = t^n$ for the differentation operator equal to 1?

Any polynomial can understood as a vector in a vector space. In this case the author implicitly chooses the monomial basis. So with the basis $(x^n)_n$ the vector $x^n$ is a vector which $n+1$th component is 1 and every other component is zero. This vector trivially has norm 1. Since something like $(0, 0, .... , 0, 1, 0, 0, ...)$ represents your monomial over the monomial basis and is in fact a unit vector.

The derivative of that polynomial is $nx^{n-1}$ which is $n$ in the $n$th component and zero everywhere else. So if you were trying to pick a bound for this derivative operator working on the basis of monomial i could always find a polynomial of sufficiently large degree to violate your bound.

If you were to choose a different basis such as $(\frac{x^n}{n!})_n$ this would no longer hold. However the linear operator which changes basis from $(\frac{x^n}{n!})_n$ to $(x^n)_n$ would be unbounded instead.