I have an identity that expresses the $n$ times integrated Brownian motion and I would like to prove that. First, I define what I mean by $n$ times integrated Brownian motion. $$V_1(t) = \int_0^tB_s\, ds \qquad V_2(t) = \int_0^tV_1(s)\, ds$$ In general, $$V_n(t) = \int_0^tV_{n-1}(s)ds$$ The identity that needs proof is as follows. $$V_n(t) = \frac{1}{n!}\int_0^t(t-s)^n\,dB_s$$
Proof by induction seems appropriate here. By using the following I show that the identity holds when $n=1$. $$tB_t = \int_0^tB_s\,ds + \int_0^ts\,dB_s$$
Then I assume that it holds for some arbitrary $n \geq 1$ and I try to show that the identity holding for $n+1$ is implied.
$$V_{n+1}(t) = \int_0^tV_n(s)\,ds = \frac{1}{n!}\int_0^t\int_0^s(s-r)^n\,dB_r\,ds$$
Next I apply Ito's lemma to the function $f(t,B_t) = (s-t)^nB_t$ for some $0<s<\infty$. This gives
$$(s-t)^nB_t = -n\int_0^t(s-r)^{n-1}B_r\,dr+\int_0^t(s-r)^n\,dB_r$$ I set $t = s$ and I substitute the resulting expression for $\int_0^s(s-r)^n\,dB_r$ to the expression for $V_{n+1}(t)$. $$V_{n+1}(t) = \frac{1}{(n-1)!}\int_0^t\int_0^s(s-r)^{n-1}B_r\,dr\,ds$$
I need to show that $$\frac{1}{(n-1)!}\int_0^t\int_0^s(s-r)^{n-1}B_r\,dr\,ds = \frac{1}{(n+1)!}\int_0^t(t-s)^{n+1}dB_s$$ Manipulating the RHS using a similar method as above I get $$n\int_0^t\int_0^s(s-r)^{n-1}B_r\,dr\,ds = \int_0^t(t-s)^{n}B_s\,ds$$ I cannot show why this is true. I am also not sure if my approach is the right one here. Could someone help please?
By Fubini's theorem,
$$\begin{align*} \int_0^t \int_0^s (s-r)^{n-1} B_r \, dr \, ds &= \int_0^t \left(\int_r^t (s-r)^{n-1} \, ds\right) B_r \, dr\\ &= \frac{1}{n} \int_0^t (t-r)^n \, B_r \, dr. \end{align*}$$
Combining this with the last equation in your question, finishes the proof.
Remark: Actually, we only have to do the "manipulation" using Itô's formula in order to switch both integrals. Statements as the stochastic Fubini show that this is also possible for stochastic integrals, i.e. it follows directly (e.g. from the stochastic Fubini) that
$$\int_0^t \int_0^s (s-r)^n \, dB_r \, ds = \int_0^t \int_r^t (s-r)^n \, ds \, dB_r.$$
This simplifies calculations a lot.