Given a standard one dimensional Brownian motion (starting in 0) $(B_t)_{t \geq 0}$ I am supposed to proof that $B_t$, $B_t^2 - t$ and $B_t^3 - 3tB_t$ are martingales, which I was able to do without a problem.
When mentioning this exercise in the lecture, my professor mentioned Hermite polynomials. So, in addition to proofing the martingales from above, for each Hermite polynomial $H_n = (-1)^n e^{x^2/2} \frac{d^n}{dx^n}e^{-x^2/2}$ I wanted to proof (if even correct) that $M^n_t := \sqrt{t^n}H_n(\frac{W_t}{\sqrt{t}})$ are all martingales. For $n=1,2,3$ this corresponds to the martingales from above.
I tried proofing this fact using direct computation: $$ \mathbb{E}[M^n_t|\mathcal{F}_s] = \mathbb{E}\left[\sqrt{t^n}H_n(\frac{B_t}{\sqrt{t}})|\mathcal{F}_s\right] = \mathbb{E}\left[\sqrt{t^n}H_n(\frac{B_t-B_s+B_s}{\sqrt{t}})|\mathcal{F}_s\right] $$ Since the increment $B_t-B_s$ is independent from $\mathcal{F}_s$ and $B_s$ is $\mathcal{F}_s$-measurable, we have $$ = \mathbb{E}\left[\sqrt{t^n}H_n(\frac{B_t-B_s+x}{\sqrt{t}})\right]\Bigg|_{x=W_s}. $$ The input from $H_n$ from the last term is normally distributed and I tried to calculate the expected value using the normal density, but I was not able to get anywhere with this.
Is this the right or is there a different, more effective way to proof that $M^n$ are martingales?
Edit: Using the comment from Rhys Steele I got a little closer. Unfortunately I can not use the Ito formula yet. But I have the following: $$ S_t:= e^{\lambda B_t - \frac12\lambda^2t} = \sum_{n=0}^\infty \frac{\lambda^n}{n!}t^{n/2}H_n(\frac{B_t}{\sqrt{t}}). $$ Additionally I know that $S_t$ is a martingale. So I have to argue that $$ S_s = \mathbb{E}[S_t|\mathcal{F}_s] = \mathbb{E}[\sum_{n=0}^\infty \frac{\lambda^n}{n!}t^{n/2}H_n(\frac{B_t}{\sqrt{t}})|\mathcal{F}_s] \overset{\textbf{???}}{=} \sum_{n=0}^\infty \frac{\lambda^n}{n!}t^{n/2}\mathbb{E}[H_n(\frac{B_t}{\sqrt{t}})|\mathcal{F}_s] $$ Then I can compare coefficients and I am done. But I am stuck an whether I can exchange the series and the expected value. I have boiled this down to show that for $$ \mathbb{E}[|H_n(\frac{B_t}{\sqrt{t}})|] = b_n $$ I still have $$ \sum_{n=0}^\infty \frac{\lambda^n}{n!}t^{n/2}b_n < \infty. $$ Does someone know a way to bound (depending on $n$) the $b_n$ to show the last inequality?