Show that the following equivalence of the $n$th stochastic integral
$$n!\int\dots\int_{0\leq s_1\dots\leq s_n\leq t}dW_{s_1}\dots dW_{s_n}=t^{n/2}H_n \left(\frac{W_t}{\sqrt{t}} \right)$$
where $H_n$ is the Hermite polynomial.
We know that the Hermite polynomial is:
$$H_n(x)=(-1)^n e^{\frac{x^2}{2}}(\frac{d^n}{dx^n}e^{-\frac{x^2}{2}}),$$ moreover we have the recursive relation $$H_{n+1}(x)=2xH_{n}(x)-H_{n}'(x).$$
If we consider the case $n=2$ the stochastic integral:
$$\begin{align*} 2\int_{0}^{s_1}\int_{s_1}^tdW_{u} \, dW_{s_2} &=2\int_{0}^{s_1}\left(\int_{s_1}^tdW_{u} \right)dW_{s_2} \\& =2\int_{0}^{s_1}(W_t-W_{s_1})\, dW_{s_2} \\ & =2 \left(\int_{0}^{s_1}W_t \, dW_{s_2}-\int_{0}^{s_1}W_{s_1} \, dW_{s_2} \right) \\ &=2 \left(\int_{0}^{s_1}W_t \, dW_{s_2}-\int_{0}^{s_1}W_{s_1} \, dW_{s_2} \right) \end{align*}$$
where we have $\int_{0}^{s_1}W_{s_1}dW_{s_2}=\frac{W_{s_1}-s_1}{2}$. On the other hand, we have:
$$H_2 \left(\frac{W_t}{\sqrt{t}} \right)=4\frac{W_t^2}{t}-2 \quad \implies \quad tH_2 \left(\frac{W_t}{\sqrt{t}} \right)=2(W_t^2-t)$$
How can I generalise it?
If we set
$$F_n(t,x) := \frac{(-t)^n}{n!} \exp \left( \frac{x^2}{2t} \right) \frac{d^n}{dx^n} \exp \left(- \frac{x^2}{2t} \right),$$
then a straight-forward application of the chain rule shows that
$$F_n(t,x) = \frac{1}{n!} t^{n/2} H_n \left( \frac{x}{\sqrt{t}} \right),$$
and therefore the assertion is equivalent to
$$\int \ldots \int_{0 \leq s_1 \leq \ldots \leq s_n \leq t} dW_{s_1} \ldots \, dW_{s_n} = F_n(t,W_t). \tag{1}$$
An application of Itô's formula and elementary (but somewhat messy) computations show that
$$dF_{n+1}(t,W_t) = F_n(t,W_t) \, dW_t \tag{2},$$
see this answer for details, and then $(1)$ follows by a simple induction. (Mind that the notation which is used in the linked answer is different from the notation in the present question/answer!)