Define the Volterra Operator $V:C([0,1])\to C([0,1])$ given by $$ Vf(x) = \int_0^x f(t) \, dt, \quad f \in C([0,1]). $$ I'm asked to proof that $\|V^n\| = \frac{1}{(n-1)!}$.
My attempt: First, I'm gonna proof that $V^n f (x) = \frac{1}{(n-1)!} \, \int_0^x (x-t)^{n-1} f(t) \, dt \quad (I)$. Which holds for $n=1$, assume that $(I)$ holds for some $n > 1$. Then,
$$ V^{n+1} f(x) = \int_0^x V^n f(t) dt = \int_0^x \left [ \frac{1}{(n-1)!}\int_0^t (t-s)^{n-1} f(s) ds \right ] dt =\\ \frac{1}{(n-1)!} \int_0^x f(s)\left [\int_s^x (t-s)^{n-1} \right ] ds = \frac{1}{n!} \int_0^x (x-s)^n f(s) ds.$$
$$|V^n f (x)| \leq \frac{1}{(n-1)!} \, \int_0^x |x-t|^{n-1} |f(t)| \, dt \leq \frac{\|f\|_0}{(n-1)!} \, \int_0^x (x-t)^{n-1} \, dt = \frac{\|f\|_0}{n!} x^n $$
Hence, $\|V^n f \| \leq \frac{1}{n!} \, \|f\|_0 \Rightarrow \|V^n\| \leq \frac{1}{n!}$ and $V^n 1 (x) = \frac{x^n}{n!}$, i.e $\| V^n 1 \|_0 = \frac{1}{n!}$.
With this I conclude that $\|V^n \| = \frac{1}{n!}$ and not $\frac{1}{(n-1)!}$.
What Am I doing wrong in this exercise?
