Show that for some Brownian Motion $B_t$, that the following is a martingale $$Y_t = \frac{1}{\sqrt{1+t}}e^{\frac{B_t^2(1+t)}{2}}$$
I want to therefore show that $\mathbb{E}(Y_t|\mathcal{F}_s) = Y_s. $
\begin{align} E[Y_t|\mathcal F_s]&= \frac{1}{2\sqrt{1+t}}E[e^{B_t^2(1+t)}|\mathcal F_s] \\ \end{align} Then consider \begin{align} E[e^{cB_t^2}|\mathcal F_s]&=E[\exp(c(B_s + B_t - B_s)^2)|\mathcal F_s]\\ &= E[\exp\left({cB_s}^2 + 2cB_s(B_t - B_s) + c(B_t - B_s)^2\right)|\mathcal F_s] \\& = E[e^{c{B_s}^2}|\mathcal F_s] \times E[\exp(2cB_s(B_t - B_s))|\mathcal F_s] \times E[e^{c(B_t - B_s)^2}|\mathcal F_s] \\&= e^{c{B_s}^2}\times E[\exp(2cB_s(B_t - B_s))|\mathcal F_s] \times e^{\frac{c(t-s)^4}{4}} \end{align}