Consider a Poisson process $\{N_t, F_t\}_{0 \le t < \infty}$. Then the martingale $M_t = N_t - \lambda t$ has quadratic variation $\lambda t$.
The quadratic variation of $M$ is defined here as the unique natural increasing process in the Doob-Meyer decomposition of $M^2$. So it suffices to show that $M_t^2 - \lambda t$ is a martingale.
Fix $s<t$. We have $\mathbb{E}[N_t -\lambda t | F_s] = N_s - \lambda s$, so $\mathbb{E}[N_t | F_s] = N_s - \lambda s + \lambda t.$ Using this, and the independence of $N_t - N_s$ to the filtration $F_s$, we have to show that $$ \mathbb{E}[N_t^2 - 2\lambda t N_t + \lambda^2 t^2 | F_s] = N_s^2 -2\lambda s N_s + \lambda^2 s^2. $$
First, $$ \begin{split} \mathbb{E}[N_t^2 |F_s] &= \mathbb{E}[(N_t-N_s+N_s)^2|F_s] \\ &= \mathbb{E}[(N_t-N_s)^2|F_s] + 2\mathbb{E}[(N_t-N_s)N_s|F_s] + N_s^2 \\ &= \mathbb{E}[(N_t-N_s)^2] + 2N_s \mathbb{E}[N_t - N_s]+N_s^2 \\ &= 2\lambda(t-s) + 2\lambda(t-s) N_s + N_s^2. \end{split} $$
Using this, I get $$ \begin{split} \mathbb{E}[N_t^2 - 2\lambda t N_t + \lambda^2 t^2 | F_s] &= N_s^2 + 2\lambda(t-s)N_s + 2\lambda(t-s) -2\lambda t(N_s - \lambda s + \lambda t) +\lambda^2t^2 \\ &= N_s^2 -2\lambda sN_s + 2\lambda t - 2 \lambda s + 2 \lambda^2 ts - \lambda^2 t^2. \end{split} $$
So I don't get the desired identity. What am I doing wrong here?
Since the expectation and the variance of $N_t-N_s$ are both equal to $\lambda(t-s)$ and since the increments of $N$ are independent we have $$ \mathbb{E}[(N_t-N_s)^2|{\cal F}_s]=\mathbb{E}[(N_t-N_s)^2]=\underbrace{\lambda(t-s)}_{\text{variance}}+\underbrace{\lambda^2(t-s)^2}_{\big(\mathbb E[N_t-N_s]\big)^2}\,.\tag{1} $$
Using (1) and \begin{align} \mathbb E[N_t|{\cal F}_s]&=\mathbb E[N_t-\lambda t|{\cal F}_s]+\lambda t=N_s-\lambda s+\lambda t\,,\tag{2} \end{align} we get \begin{align} \\[2mm] \mathbb{E}[N_t^2 |{\cal F}_s] &= \mathbb{E}[(N_t-N_s+N_s)^2|{\cal F}_s] \\[2mm] &= \mathbb{E}[(N_t-N_s)^2|{\cal F}_s] + 2\,\mathbb{E}[(N_t-N_s)N_s|{\cal F}_s] + N_s^2 \\[2mm] &= \mathbb{E}[(N_t-N_s)^2] + 2\,N_s\,\mathbb{E}[N_t - N_s]+N_s^2 \\[2mm] &= \underbrace{\lambda(t-s)+\lambda^2(t-s)^2}_{(1)} + 2\lambda(t-s) N_s + N_s^2\,\tag{3} \end{align} and
\begin{align}\require{cancel} &\mathbb E[(N_t-\lambda t)^2|{\cal F}_s]=\mathbb E[N_t^2-2\lambda tN_t+\lambda^2t^2|{\cal F}_s]\\[2mm] &=\underbrace{\lambda(t-s)+\lambda^2(t-s)^2 + 2\lambda(t-s) N_s + N_s^2}_{(3)}-2\lambda t\underbrace{\big(N_s-\lambda s+\lambda t\big)}_{(2)}+\lambda^2t^2\\[2mm] &=\lambda(t-s)+\lambda^2(t-s)^2 + 2\lambda(\bcancel{t}-s) N_s + N_s^2-\bcancel{2\lambda tN_s}-2\lambda^2t(t-s)+\lambda^2t^2\\[2mm] &=\lambda(t-s)+\lambda^2(t-s)^2 \underbrace{- 2\lambda s N_s + N_s^2}_{(*)}-2\lambda^2t(t-s)+\lambda^2t^2\\[2mm] &=\lambda(t-s)+\lambda^2(\cancel{t^2}-\bcancel{2ts}+\xcancel{s^2})+\overbrace{(N_s-\lambda s)^2-\xcancel{\lambda^2s^2}}^{(*)} -2\lambda^2(\cancel{t^2}-\bcancel{ts})+\cancel{\lambda^2t^2}\\[2mm] &=\lambda(t-s)+(N_s-\lambda s)^2\,. \end{align} This shows that $(N_t-\lambda t)^2-\lambda t$ is a martingale and therefore $\langle N\rangle_t=\lambda t\,.$