Showing $e^{2B_t - 2t}$ is a martingale

Question

A process $M_t$ is a martingale, if

$1(a): \space \space \mathbb{E}[M_t | \mathcal{F}_s] = M_s$ for all $s \leq t$

Or equivalently,

$1(b): \space \space \mathbb{E}[M_{t+s}| \mathcal{F}_t] = M_t$ for all $s \leq t$

I can't seem to solve this problem, using the notation in definition 1(a).

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I start:

$$\mathbb{E}[e^{2B_t - 2t} | \mathcal{F}_s] = e^{-2t}\mathbb{E}[e^{2(B_t - B_s + Bs}| \mathcal{F}_s]$$

is this first step correct?

If I keep going, I end up with:

$$e^{2(B_s - t)}\mathbb{E}[e^{2(B_t - B_s)}] = e^{2(B_s - t)}e^{2(t-s)^2}$$

Which isn't $M_s$

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I think I make a mistake when I use the properties of MGF.

Is it correct to say: $$2(B_t - B_s) \sim N(0, 4(t-s))$$

$$\mathbb{E}[e^{2(B_t - B_s}] = e^{(2(t-s)^2)}$$ using the fact that

$$\mathbb{E}[e^{\alpha X}] = e^{\alpha \mu + \frac{\sigma ^2 \alpha ^2}{2}}$$

Answer 1

$$\mathbb{E}[e^{\alpha X}] = e^{\alpha \mu + \frac{\sigma ^2 \alpha ^2}{2}} = e^{2 (0) + \frac{(t-s) 2^2}{2}} = e^{2(t-s)}$$