Why is $e^{W_t}$ not a martingale? Any advice on the following will be appreciated. 
I am trying to show that for $X_t = e^{W_t}$, $X_t$ is not a martingale, where $W_t$ is Brownian motion with respect to filtration $\mathcal F_t$
My approach is $\forall s < t$:
\begin{align}
E[X_t|Fs] & = E[e^{W_t}|\mathcal F_s] \\
&= E[e^{W_t}] \\ 
& = e^{t/2}
\end{align}


*

*Is $e^{W_t}$ be independent to $\mathcal F_s$?


I used a pre-established result for the second to third line. 
The following is another approach, which gives a different answer,
\begin{align}
E[X_t|\mathcal F_s] &= E[e^{W_t}|\mathcal F_s] \\
&= E[e^{W_t - W_s + W_s}|\mathcal F_s] \\ 
&= e^{W_s}E[e^{W_t - W_s}|\mathcal F_s] \\
&=e^{W_s}e^{\frac{t-s}{2}}
\end{align}
I used the property $E[e^{W_t - W_s}|\mathcal F_s] = e^{\frac{t-s}{2}}$
Two different approaches that led to two different answers. I believe the former is wrong, but I cannot tell why it is wrong.
Thank you! 
 A: The mistake in the first approach occurred in the second equality. It is not true (when $s<t$) that $\mathbb E[e^{W_t}\mid \mathcal F_s]=e^{W_t}$. In fact, the simplest way I see to compute this conditional expectation (for $s<t$) is to use the fact that $W_t$ is a Markov process, and therefore so is $e^{W_t}$. Thus,
$$
\mathbb E[e^{W_t}\mid \mathcal F_s]=\mathbb E[e^{W_t}\mid W_s]=e^{W_s}\mathbb E[e^{W_t-W_s}\mid W_s].
$$
Now use the independence of increments to conclude that 
$$
E[e^{W_t-W_s}\mid W_s]=\mathbb Ee^{W_t-W_s}=e^{(t-s)/2}.
$$
Consequently,
$$
\mathbb E[e^{W_t}\mid \mathcal F_s]=e^{(t-s)/2}W_s.
$$
In summary, your second approach is correct, although it sounds like it is a transcription of an argument you heard somewhere and didn't fully understand. Thus in my answer I have attempted to walk through the same steps but in more detail. Please ask if there is a specific step that doesn't make sense or that you are treating as a "black box" without understanding why it is true.
