$M_t = \frac{1}{\sqrt{T_1}} \mathbb{1} (T_1 \leq t) - 2 \sqrt{T_1 \wedge t}, t\geq 0$ is a martingale I try to solve an old exam question, but I find it difficult. Maybe someone can suggest a hint.
Let $\{ T_i | i\in \mathbb{N} \}\subseteq \mathbb{R} _{\geq 0}$ be a homogeneous Poission point process with intensity 1, where we assume $T_1 < T_2 < \dots$. Let $M_t = \frac{1}{\sqrt{T_1}} \mathbf{1} (T_1 \leq t) - 2 \sqrt{T_1 \wedge t}$. Let $N_t = \# \{ i\in \mathbb{N} | T_i \leq t \}$. Let $\mathcal{F} _t = \sigma ( (N_s) _{s\leq t} )$. Prove that $(M_t)$ is an $(\mathcal{F} _t)$-martingale.
I already showed that $M_t$ is $\mathcal{F} _t$-adapted by taking a careful look at events of the form $\{  \frac{1}{\sqrt{T_1}} \mathbf{1} (T_1 \leq t) \leq x \}$ and $\{ \sqrt{T_1 \wedge t} \leq x \}$. I proved integrability by using upper bounds and the integration by parts technique. Now I have to show that $M_t$ satisfies the martingale property. So let $s<t$. Then:
\begin{align*}
E [ M_t | \mathcal{F} _s ] & = E \left [\frac{1}{\sqrt{T_1}} \mathbf{1} (T_1 \leq t) - 2 \sqrt{T_1 \wedge t} \ \mid \ \mathcal{F}_s \right ]\\
& = E \left [\frac{1}{\sqrt{T_1}} \mathbf{1} (T_1 \leq t) \ \mid \ \mathcal{F}_s \right ] - 2  E \left [ \sqrt{T_1 \wedge t} \ \mid \ \mathcal{F}_s \right ] \\
 & = E \left [\frac{1}{\sqrt{T_1}} (\mathbf{1} (T_1 \leq s) + \mathbf{1} (s < T_1 \leq t)) \ \mid \ \mathcal{F}_s \right ] - 2  E \left [ \sqrt{T_1 \wedge t} \ \mid \ \mathcal{F}_s \right ] \\
& = E \left [\frac{1}{\sqrt{T_1}} (\mathbf{1} (N_s \geq 1) + \mathbf{1} (N_s = 0) \cdot \mathbf{1} (N_t - N_s \geq 1))\ \mid \ \mathcal{F}_s \right ] - 2  E \left [ \sqrt{T_1 \wedge t} \ \mid \ \mathcal{F}_s \right ] \\
& = \frac{1}{\sqrt{T_1}} \mathbf{1} (N_s \geq 1) + \frac{1}{\sqrt{T_1}} \mathbf{1} (N_s = 0) \cdot P (N_t - N_s \geq 1)\  - 2  E \left [ \sqrt{T_1 \wedge t} \ \mid \ \mathcal{F}_s \right ] \\
& = \frac{1}{\sqrt{T_1}} \mathbf{1} (N_s \geq 1) + \frac{1}{\sqrt{T_1}} \mathbf{1} (N_s = 0) \cdot (1-\exp(s-t))\  - 2  E \left [ \sqrt{T_1 \wedge t} \ \mid \ \mathcal{F}_s \right ] \\
& = \frac{1}{\sqrt{T_1}} \mathbf{1} (T_1 \leq s) + \frac{1}{\sqrt{T_1}} \mathbf{1} (T_1 > s) \cdot (1-\exp(s-t))\  - 2  E \left [ \sqrt{T_1 \wedge t} \ \mid \ \mathcal{F}_s \right ] \\
& = \frac{1}{\sqrt{T_1}} \mathbf{1} (T_1 \leq s) + \frac{1}{\sqrt{T_1}} \mathbf{1} (T_1 > s) \cdot (1-\exp(s-t))\  - 2  E \left [ \sqrt{T_1} \cdot \mathbf{1} (T_1 \leq s) + \sqrt{T_1} \cdot \mathbf{1} (s < T_1 \leq t) + \sqrt{t} \cdot \mathbf{1} (T_1 > t) \ \mid \ \mathcal{F}_s \right ] \\
& = \frac{1}{\sqrt{T_1}} \mathbf{1} (T_1 \leq s) + \frac{1}{\sqrt{T_1}} \mathbf{1} (T_1 > s) \cdot (1-\exp(s-t))\  - 2  E \left [ \sqrt{T_1} \cdot \mathbf{1} (N_s \geq 1) + \sqrt{T_1} \cdot \mathbf{1} (N_s = 0) \cdot \mathbf{1} (N_t - N_s \geq 1) + \sqrt{t} \cdot \mathbf{1} (N_t = 0) \ \mid \ \mathcal{F}_s \right ] \\
& = \frac{1}{\sqrt{T_1}} \mathbf{1} (T_1 \leq s) + \frac{1}{\sqrt{T_1}} \mathbf{1} (T_1 > s) \cdot (1-\exp(s-t))\  - 2  \sqrt{T_1} \cdot \mathbf{1} (N_s \geq 1) - 2\sqrt{T_1} \cdot \mathbf{1} (N_s = 0) \cdot P (N_t - N_s \geq 1) - 2 \sqrt{t} \cdot P (N_t = 0)  \\
& = \frac{1}{\sqrt{T_1}} \mathbf{1} (T_1 \leq s) + \frac{1}{\sqrt{T_1}} \mathbf{1} (T_1 > s) \cdot (1-\exp(s-t))\  - 2  \sqrt{T_1} \cdot \mathbf{1} (N_s \geq 1) - 2\sqrt{T_1} \cdot \mathbf{1} (N_s = 0) \cdot (1-\exp(s-t)) - 2 \sqrt{t} \cdot \exp(-t)  \\
& = \frac{1}{\sqrt{T_1}} \mathbf{1} (T_1 \leq s) + \frac{1}{\sqrt{T_1}} \mathbf{1} (T_1 > s) \cdot (1-\exp(s-t))\  - 2  \sqrt{T_1} \cdot \mathbf{1} (T_1 \leq s) - 2\sqrt{T_1} \cdot \mathbf{1} (T_1 > s) \cdot (1-\exp(s-t)) - 2 \sqrt{t} \cdot \exp(-t)  \\
\end{align*}
I do not know how to proceed further? Can someone give a hint? Thank you in advance.
EDIT: I think I need to use the memoryless property somewhere. Also, I possibly made one or more mistakes with $\mathcal{F} _s$-measurability and conditioning on $\mathcal{F}_s$.

Note that we can write
$$ 2\sqrt{T_1 \wedge t} = \int_{0}^{t} \frac{1}{\sqrt{u}} \mathbf{1}(T_1 > u) \, \mathrm{d}u. $$
From this, for $0 \leq s \leq t$,
\begin{align*}
\mathbb{E}[ M_t - M_s \mid \mathcal{F}_s]
&= \mathbb{E}\biggl[ \frac{1}{\sqrt{T_1}} \mathbf{1}(s < T_1 \leq t) \,\biggm|\, \mathcal{F}_s \biggr] - \int_{s}^{t} \frac{1}{\sqrt{u}} \mathbb{E}[ \mathbf{1}(u < T_1) \mid \mathcal{F}_s] \, \mathrm{d}u \tag{*}
\end{align*}
The memoryless property tells that, given $\{T_1 > s\}$, the extra waiting time $T_1 - s$ is also $\operatorname{Exp}(1)$ and independent of $\mathcal{F}_s$. From this, the right-hand side of $\text{(*)}$ simplifies to
\begin{align*}
&\biggl( \mathbb{E}\biggl[ \frac{1}{\sqrt{s + T_1}} \mathbf{1}(T_1 \leq t-s) \biggr] - \int_{s}^{t} \frac{1}{\sqrt{u}} \mathbb{P}(u-s < T_1) \, \mathrm{d}u \biggr) \mathbf{1}(s < T_1) \\
&= \biggl( \int_{0}^{t-s} \frac{1}{\sqrt{s+u}} e^{-u} \, \mathrm{d}u - \int_{s}^{t} \frac{1}{\sqrt{u}} e^{-(u-s)} \, \mathrm{d}u \biggr) \mathbf{1}(s < T_1).
\end{align*}
It is easy to verify that the two integrals in the last line are equal, hence the desired conclusion follows.
