Convergence in Distribution of the maximum of a sequence.

Unfortunately we do not have convergence in probability, so that approach is not going to succeed. The basic trick is to note that $M_n\le a$ if and only if $X_i\le a$ for each $i=1,\ldots,n$ and then use independence: $$P(M_n-\tfrac1\lambda\log n\le x)=P(M_n\le\tfrac1\lambda\log n+x)=\prod_{i=1}^nP(X_i\le\tfrac1\lambda\log n+x).$$ Now remembering that $P(X_i\le a)=1-e^{-\lambda a}$, we find $$P(M_n-\tfrac1\lambda\log n\le x)=\left(1-e^{-\lambda(\frac1\lambda\log n+x)}\right)^n=\left(1-\frac{e^{-\lambda x}}n\right)^n\to e^{-e^{-\lambda x}}$$ as $n\to\infty$. Hence $M_n\to Z$ in distribution where $P(Z\le x)=e^{-e^{-\lambda x}}$.