Contraction semigroup on $X$ with generator $A$ Let $\{S(t)\}_{t \ge 0}$ be a contraction semigroup on $X$, with generator $A$. Inductively define $D(A^k):=\{u\in D(A^{k-1}) \mid A^{k-1}u \in D(A)\}$ ($k=2,\ldots$). Show that if $u \in D(A^k)$ for some $k$, then $S(t)u \in D(A^k)$ for each $t \ge 0$.

This is from PDE Evans, 2nd edition: Chapter 7, Exerise 15.
For reference, here is Theorem 1. This is on page 435 in the textbook.

THEOREM 1 (Differential properties of semigroups). Assume $u \in D(A)$. Then
(i) $S(t)u\in D(A)$ for each $t \ge 0$
(ii) $AS(t)u=S(t)Au$ for each $t \ge 0$.
(iii) The mapping $t \mapsto S(t)u$ is differentiable for each $t&gt;0$.
(iv) $\frac d{dt}S(t)u = AS(t)u \quad (t &gt; 0)$.

May I receive feedback on my following proof to this problem?
My attempted proof:
If $u \in D(A^k)$, then $u \in D(A^{k-1})$ and $A^{k-1}u \in D(A)$. If $u \in D(A^{k-1})$, then $u \in D(A^{k-2})$ and $A^{k-2}u \in D(A)$. Inductively proceeding, we obtain: If $u \in D(A^2)$, then $u \in D(A)$ and $Au \in D(A)$. By parts (i) and (ii) of the theorem, $S(t)u \in D(A)$ and $AS(t)u=S(t)Au \in D(A)$, respectively; hence, $S(t)u \in D(A^2)$. Inductively proceeding back to $k$, we obtain: $S(t)u \in D(A^{k-1})$ and $A^{k-1}S(t)u \in D(A)$; hence, $S(t)u \in D(A^k)$.
Yes, basically that's it.

It might be only a typo, but using part (i) for $A^{k-1}u\in D(A)$ yields $S(t)A^{k-1}u\in D(A)$, and thus you need the exchange rule ($A^{k-1}S(t)u=S(t)A^{k-1}u$) to conclude $u\in D(A^k)$.

Note also that you implicitly used $S(t)A^{d}u\in D\left(A^{k-d}\right)$ for each $d=0\dots k$. This might need a proof.
Yes, basically that's it.

It might be only a typo, but using part (i) for $A^{k-1}u\in D(A)$ yields $S(t)A^{k-1}u\in D(A)$, and thus you need the exchange rule ($A^{k-1}S(t)u=S(t)A^{k-1}u$) to conclude $u\in D(A^k)$.

Note also that you implicitly used $S(t)A^{d}u\in D\left(A^{k-d}\right)$ for each $d=0\dots k$. This might need a proof.