The <Laplace-transform formula for a semigroup resolvent> is locally uniformly convergent in the half-plane $\operatorname{Re}z>\omega$, so it may be differentiated under the <Bochner integral>:
$$
\frac{d^n}{dz^n}R(z)u
=(-1)^n\int_0^\infty t^ne^{-tz}U(t)u\,dt.
$$
The <resolvent identity> gives $R'(z)=-R(z)^2$ and, by <mathematical induction>,
$$
\frac{d^n}{dz^n}R(z)=(-1)^nn!R(z)^{n+1}.
$$
Therefore
$$
\boxed{n!(zI-A)^{-(n+1)}u
=\int_0^\infty t^ne^{-tz}U(t)u\,dt}.
$$
For real $\lambda>\omega$, the <integral triangle inequality> and the <Gamma integral> give
$$
\begin{aligned}
n!\|(\lambda I-A)^{-(n+1)}u\|
&\leq M\|u\|\int_0^\infty t^ne^{-(\lambda-\omega)t}\,dt\\
&=\frac{Mn!}{(\lambda-\omega)^{n+1}}\|u\|.
\end{aligned}
$$
Consequently
$$
\boxed{\|(\lambda I-A)^{-(n+1)}u\|
\leq\frac{M}{(\lambda-\omega)^{n+1}}\|u\|}.
$$
The exponent $-n+1$ printed in the question is a typographical error: already at $n=0$ it contradicts the first-resolvent bound.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 319 / 1 / b / Solution

For two points $z,w$ in the resolvent set of an operator $A$,
$$
R(z,A)-R(w,A)=(w-z)R(z,A)R(w,A).
$$
In particular, the <resolvent operator> is differentiable and $\partial_zR(z,A)=-R(z,A)^2$.

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 Incoming links: Resolvent identity