Convergence of Maximum of $n$ i.i.d. random variables in probability and almost surely

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Problem: Let $\{X_k\}_{k\in\mathbb N}$ be a sequence of i.i.d. $\text{Unif}(0,1)$ random variables. Let $M_n=\max\{X_1,\dots,X_n\}$ be the maximum of the first $n$ random numbers.
$\textbf{a)}$ Show that $M_n\overset{P}{\longrightarrow}1.$
Fix $0<\varepsilon<1$. We have \begin{align*} P(\vert M_n-1\vert\geq\varepsilon)&=1-P(\vert M_n-1\vert<\varepsilon)=1-P(1-\varepsilon<M_n<1+\varepsilon)\\ &=1-\int_{1-\varepsilon}^1 nx^{n-1}\,dx=(1-\varepsilon)^n\\ &\overset{n\to\infty}\longrightarrow 0. \end{align*} For any $\delta>\varepsilon$ we have $$P(\vert M_n-1\vert\geq\delta)\leq P(\vert M_n-1\vert\geq\varepsilon)\overset{n\to\infty}\longrightarrow0.$$ It follows that $P(\vert M_n-1\vert\geq\varepsilon)\overset{n\to\infty}\longrightarrow0$ for all $\varepsilon>0$, and thus $M_n\overset{P}{\longrightarrow}1.$

$\textbf{b)}$ Show that $M_n\longrightarrow1$ almost surely.
Fix $0<\varepsilon<1$. Then we have $$\sum_{n=1}^\infty P(\vert X_n-X\vert\geq\varepsilon)=\sum_{n=1}^\infty(1-\varepsilon)^n=\frac{1-\varepsilon}{\varepsilon}<\infty,$$ where we used the formula for the sum of a geometric series, which we can do since $0<1-\varepsilon<1$. Moverover, for any $\delta>\varepsilon$ we have $$P(\vert M_n-1\vert\geq\delta)\leq P(\vert M_n-1\vert\geq\varepsilon),$$ so that we actually have for any $\varepsilon>0$, $$\sum_{n=1}^\infty P(\vert X_n-X\vert\geq\varepsilon)<\infty.$$ It follows from the Borel-Cantelli lemma that $M_n\longrightarrow1$ almost surely.


Do you agree with my reasoning above? Thank you for your time and feedback, it is much appreciated.