Consider $C[0,1]$ as a Banach space with $\lVert f \rVert_\infty = \max_{t \in [0,1]} |f(x)|$, and define $T \in B(C[0,1])$ by $Tf(t) = \int_0^t f(s) ds$
1- Show that $T^n f(t) = \frac{1}{(n-1)!} \int_0^t (t-s)^{n-1} f(s) ds$ for $n \in \mathbb{N}$.
2- Determine $\sigma(T)$. Hint: $r(T) = \underset{n \to \infty}\lim {\lVert T^n \rVert^{\frac{1}{n}}}$.
Edited: I solved the first part using the hints. Thank you so much. However, the second part is totally unclear for me, any hints please.
Using the hints mentioned above, let start by applying the operator many times to see the result.
\begin{align} T(f(t)) &= \int_0^t f(s) ds\\ T (T(f(t)))=T^2 (f(t) &= \int_0^t \left( \int_0^\tau f(\tau) d\tau \right)ds\\ &= \int_0^t \int_0^\tau f(\tau) d\tau ds \\ &= \int_0^t \int_\tau^t f(\tau) ds d\tau\\ &= \int_0^t \left( \int_\tau^t ds \right) f(\tau) d\tau\\ &= \int_0^t f(\tau) (t -\tau) d\tau \end{align}
So we can see that;
$$T^2 f(t)=\int_0^t f(s) (t -s) ds.$$ Applying the operator more times we get; $$T^3 f(t)= \frac{1}{2}\int_0^t f(s) (t -s)^2 ds.$$ and $$T^4 f(t)= \frac{1}{2 . 3}\int_0^t f(s) (t -s)^3 ds.$$
To prove the relation for all $n\in \mathbb{N}$ we use induction. Since the relation holds when $n=1$ we assume it holds when $n=k \in \mathbb{N}$,
$$T^k f(t)= \frac{1}{(k-1)!}\int_0^t f(s) (t -s)^{(k-1)} ds.$$
Now we apply by the operator one more time, \begin{align} T(T^k f(t)) =& \int_0^t \left( \frac{1}{(k-1)!}\int_0^s f(\tau) (s -\tau)^{(k-1)} d\tau \right) ds\\ T^{k+1} f(t) =& \frac{1}{(k-1)!} \int_0^t \int_0^s f(\tau) (s -\tau)^{(k-1)} d\tau ds\\ =& \frac{1}{(k-1)!} \int_0^t \int_\tau^t f(\tau) (s -\tau)^{(k-1)} ds d\tau\\ =& \frac{1}{(k-1)!} \int_0^t \left( \int_\tau^t ds \right) f(\tau) (s -\tau)^{(k-1)} d\tau\\ =& \frac{1}{k!} \int_0^t f(\tau) (s -\tau)^{k} d\tau.\\ \end{align}
So the relation $$T^n f(t) = \frac{1}{(n-1)!} \int_0^t (t-s)^{n-1} f(s) ds$$ is right for all $n \in \mathbb{N}$.