Knowing that M and N are two local martingales. Can I do the following simple reasoning for claiming that $N^\tau(M^\tau - M)$ is a local martingale?
: Since a stopped local martingale is still a local martingale, and since the product of two local martingale is still a local martingale, then $N^\tau M^\tau$. Applying the same reasoning also at $N^\tau M$ I could conclude that it is also a local martingale. Finally the difference of 2 local martingale is still a local martingale.
It follows that $N^\tau(M^\tau - M)$ is a local martingale.
If this reasoning does not hold, what is wrong? How could I prove it without using Ito or advanced techniques?
Let $M^{(n)}$ and $N^{(n)}$ denote $M$ and $N$ stopped at $\sigma_n\wedge\tau_n$. These are both bounded martingales. Form $X^{(n)}_t:=N^{(n)}_{t\wedge\tau}(M^{(n)}_{t\wedge\tau}-M^{(n)}_t)$. As previously noted (now corrected), $$ X^{(n)}_t=\cases{0,&$0\le t<\tau$,\cr N^{(n)}_\tau(M^{(n)}_{\tau}-M^{(n)}_t),&$t\ge \tau$.\cr} $$ Now decompose (for $0\le s<t$) $$ X^{(n)}_t=(1_{\{s<t\le\tau\}}+1_{\{s<\tau\le t\}} +1_{\{\tau\le s<t\}})\cdot X^{(n)}_t $$ into three terms. The conditional expectation of the first term (given $\mathcal F_s$) is trivial; for the second use the tower property of conditional expectations, conditioning first on $\mathcal F_\tau$; the third is easy because $M^{(n)}$ is a martingale and $N^{(n)}_\tau\cdot 1_{\{\tau\le s<t\}}$ is $\mathcal F_s$-measurable. It follows that $X^{(n)}$ is a martingale (for each $n$), and then that $X$ is a local martingale.