Can we calculate the multiple stochastic integral? Let $(W_{t})_{t \geq 0}$ ba a $d$-dimensional Brownian motion, $k \in {\mathbb N}$ and $i_{1},i_{2},\ldots,i_{k} \in \{1,2,\ldots,d\}$.
Then can we calculate the following multiple stochastic integral?
$$
{\mathbb E}\left[\prod_{j=1}^{k}\int_{0}^{t}f_{i_{j}}(s){\rm d}W_{s}^{i_{j}}\right]=???
$$
For example, for $k=1$ and $k=2$,
$$
{\mathbb E}\left[\int_{0}^{t}f_{i_{1}}(s){\rm d}W_{s}^{i_{1}}\right]=0, \quad 
{\mathbb E}\left[\prod_{j=1}^{2}\int_{0}^{t}f_{i_{j}}(s){\rm d}W_{s}^{i_{j}}\right]={\mathbb E}\left[\int_{0}^{t}f_{i_{1}}(s)f_{i_{2}}(s){\rm d}s\right]\delta_{i_{1},i_{2}}.
$$
 A: For $k=3$, using the It\^o formula, we obtain
\begin{align*}
\prod_{j=1}^{3}\int_{0}^{t}f_{i_{j}}(s){\rm d}W_{s}^{i_{j}}
&=\int_{0}^{t}\left(\int_{0}^{s}f_{i_{2}}(u){\rm d}W_{u}^{i_{2}}\right)\left(\int_{0}^{s}f_{i_{3}}(u){\rm d}W_{u}^{i_{3}}\right)f_{i_{1}}(s){\rm d}W_{s}^{i_{1}} \\
&\hspace{0.5cm}+\int_{0}^{t}\left(\int_{0}^{s}f_{i_{1}}(u){\rm d}W_{u}^{i_{1}}\right)\left(\int_{0}^{s}f_{i_{3}}(u){\rm d}W_{u}^{i_{3}}\right)f_{i_{2}}(s){\rm d}W_{s}^{i_{2}} \\
&\hspace{0.5cm}+\int_{0}^{t}\left(\int_{0}^{s}f_{i_{1}}(u){\rm d}W_{u}^{i_{1}}\right)\left(\int_{0}^{s}f_{i_{2}}(u){\rm d}W_{u}^{i_{2}}\right)f_{i_{3}}(s){\rm d}W_{s}^{i_{3}} \\
&\hspace{0.5cm}+\frac{1}{2}\int_{0}^{t}\int_{0}^{s}f_{i_{3}}(u){\rm d}W_{u}^{i_{3}}\delta_{i_{1},i_{2}}{\rm d}s \\
&\hspace{0.5cm}+\frac{1}{2}\int_{0}^{t}\int_{0}^{s}f_{i_{2}}(u){\rm d}W_{u}^{i_{2}}\delta_{i_{1},i_{3}}{\rm d}s \\
&\hspace{0.5cm}+\frac{1}{2}\int_{0}^{t}\int_{0}^{s}f_{i_{1}}(u){\rm d}W_{u}^{i_{1}}\delta_{i_{2},i_{3}}{\rm d}s.
\end{align*}
Thus by Fubini's theorem, we have
\begin{align*}
{\mathbb E}\left[\prod_{j=1}^{3}\int_{0}^{t}f_{i_{j}}(s){\rm d}W_{s}^{i_{j}}\right]
&=\frac{1}{2}{\mathbb E}\left[\int_{0}^{t}\int_{0}^{s}f_{i_{3}}(u){\rm d}W_{u}^{i_{3}}{\rm d}s\right]\delta_{i_{1},i_{2}} \\
&\hspace{0.5cm}+\frac{1}{2}{\mathbb E}\left[\int_{0}^{t}\int_{0}^{s}f_{i_{2}}(u){\rm d}W_{u}^{i_{2}}{\rm d}s\right]\delta_{i_{1},i_{3}} \\
&\hspace{0.5cm}+\frac{1}{2}{\mathbb E}\left[\int_{0}^{t}\int_{0}^{s}f_{i_{1}}(u){\rm d}W_{u}^{i_{1}}{\rm d}s\right]\delta_{i_{2},i_{3}} \\
&=\frac{1}{2}\int_{0}^{t}{\mathbb E}\left[\int_{0}^{s}f_{i_{3}}(u){\rm d}W_{u}^{i_{3}}\right]{\rm d}s\delta_{i_{1},i_{2}} \\
&\hspace{0.5cm}+\frac{1}{2}\int_{0}^{t}{\mathbb E}\left[\int_{0}^{s}f_{i_{2}}(u){\rm d}W_{u}^{i_{2}}\right]{\rm d}s\delta_{i_{1},i_{3}} \\
&\hspace{0.5cm}+\frac{1}{2}\int_{0}^{t}{\mathbb E}\left[\int_{0}^{s}f_{i_{1}}(u){\rm d}W_{u}^{i_{1}}\right]{\rm d}s\delta_{i_{2},i_{3}} \\
&=0.
\end{align*}
The general case follows similarly.
