Prove stochastic differential relation $(dN_t)^2 = dN_t$ I have some difficulties in showing the following problem. Any hints will be appreciated!
Show $$(dN_t)^2 = dN_t$$ for $N_t$ is the Poisson process 
I have thought of two approaches, both were met with difficulties
Approach 1: Use previously proven relations:
I have $(dM_t)^2 = dN_t$, where $M_t$ is the compensated Poisson process - $M_t = N_t - \lambda t$ and $d_t\,dM_t = 0$. Also $dM_t = dN_t -\lambda dt$ so 
\begin{align}
(dN_t)^2 &= (dM_t + \lambda dt)^2 \\
&= (dM_t)^2 +2\lambda dt\,dM_t + {\lambda}^2dt^2 \\
&= dN_t + {\lambda}^2dt^2
\end{align}
So the remaining piece for this approach is for the $\lambda^2dt^2$ term to disappear. But will it necessarily disappear?
Approach 2: By definition. 
Let $a < b$ and consider the partition $a = t_0 < t_1 < \cdots < t_{n-1} < t_n = b$ then I want to show $$\textrm{ms}-\lim_{||\Delta_n|| \to 0} \sum_{k=0}^{n-1}[\Delta N_k^2 - \Delta N_k] = 0$$ where $||\Delta_n|| = \sup_{0 \leq k \leq n-1}(t_{k+1} - t_k)$
I attempt to prove the above by showing $E[\Delta N_k^2 - \Delta N_k] \to 0$ and $\textrm{var}[\Delta N_k^2 - \Delta N_k] \to 0$
but was unsuccessful - so I am suspecting that approach 2 is off. 
Thanks! 
 A: Continuing from where you ended, the random variables $ΔN_k$ are independent, and all are Poisson distributed with parameter $λΔt_k$ so that expectation and variance are additive.
Now we know from the definition of a Poisson process that $ΔN_k=m$ with probability $e^{-λΔt_k}\frac{(λΔt_k)^m}{m!}$, so that
\begin{align}
\Bbb E[ΔN_k^2-ΔN_k] &= e^{-λΔt_k}\sum_{m=0}^\infty(m^2-m)\frac{(λΔt_k)^m}{m!}
\\
&=e^{-λΔt_k}\sum_{m=2}^\infty\frac{(λΔt_k)^m}{(m-2)!}
\\
&=(λΔt_k)^2
\\
\text{and }~~
{\rm Var}(ΔN_k^2-ΔN_k)
&=e^{-λΔt_k}\sum_{m=0}^\infty(m^2-m)^2\frac{(λΔt_k)^m}{m!} - (λΔt_k)^4
\\
&=e^{-λΔt_k}\sum_{m=2}^\infty(m^2-m)\frac{(λΔt_k)^m}{(m-2)!} - (λΔt_k)^4
\\
&\qquad\qquad\Bigl(\text{ and using }m^2-m=(m-2)(m-3)+4(m-2)+2\Bigr)
\\
&=(λΔt_k)^4+4(λΔt_k)^3+2(λΔt_k)^2 - (λΔt_k)^4
\\
&=2(λΔt_k)^2(1+2λΔt_k)
\end{align}
In both cases the sum over the full interval gives quantities that have bounds in multiples of $(b-a)\cdot \max_kΔt_k$. As these bounds go to zero with progressive refinement of the subdivision, one can conclude that the infinitesimal quantities $dN_t^2-dN_t$ are zero. This is a symbolic abbreviation for the fact that the quadratic variation of the Poisson process is the same process again.
