What does "maximum value" of a set of random variables mean? In our statistical mechanics lecture the professor said something along the lines of:

If we have some independent random variables $x_1,x_2,x_3,...,x_n$ having identical distributions:
Suppose $M_{n}=\text{max}(x_1,x_2,x_3,...)$, then we say that
  probability that $M_{n}<x$ is $\text{$\text{Pr}(M_{n}<x$)}$ (say).
In such a case
  $\text{$\text{Pr}(M_n<x$)}=\text{$\text{Pr}(x_1<x,x_2<x,x_3<x,...$)}=(\text{Pr}(x))^{n}$

Now, first of all I don't understand what he meant by $M_{n}=\text{max}(x_1,x_2,x_3,...)$. What does maximum of a set of random variables even mean? Does it refer to the random variable which can take the highest value?
Secondly I don't understand the step $\text{$\text{Pr}(M_n<x$)}=\text{$\text{Pr}(x_1<x,x_2<x,x_3<x,...$)}=(\text{Pr}(x))^{n}$
 A: The random variables $x_i$ are functions $\Omega\to\mathbb R$ where $\Omega$ is the sample space of the practicized probability space.
Then $M_n$ is also a function $\Omega\to\mathbb R$, and it is prescribed by:$$\omega\mapsto\max(x_1(\omega),\dots,x_n(\omega))$$
Note that: $$\{\omega\in\Omega\mid M_n(\omega)<x\}=\{\omega\in\Omega\mid x_1(\omega)<x\wedge\dots\wedge x_n(\omega)<x\}$$
so that also:$$\mathsf P(\{\omega\in\Omega\mid M_n(\omega)<x\})=\mathsf P(\{\omega\in\Omega\mid x_1(\omega)<x\wedge\dots\wedge x_n(\omega)<x\})$$or abbreviated:$$\mathsf P(M_n<x)=\mathsf P(x_1<x,\dots,x_n<x)$$
The RHS equals: $$\mathsf P(x_1<x)\times\cdots\times\mathsf P(x_n<x)$$because the $x_i$ are independent.
If further the $x_i$ have equal distribution and $\text{Pr}(x):=\mathsf P(x_1<x)$ then we finally arrive at:$$=\text{Pr}(x)^n$$
A: (Looking just at the case with 2 random variables, for simplicity.)  
The maximum $\max(x_1,x_2)$ is itself a random variable, similar to e.g. the sum $x_1 + x_2$.  If $x_1$, $x_2$ are dice rolls, $\max(x_1,x_2)$ will be “the higher of the two values rolled”.
If you take any point in the sample space (informally: each time you run the experiment), you get actual values for $x_1$ and $x_2$.  So then you can take the sum of these values, $x_1 + x_2$; similarly, you can take their maximum, $\max(x_1,x_2)$.
So $x_1 + x_2$ and $\max(x_1,x_2)$ are themselves random variables — they don’t have a fixed value, but they get a value each time you run the experiment — formally, they’re mappings from the sample space to values.
Re your second question: the key logical property of the “maximum” function is that for any numbers $a_1, \ldots, a_n$,and $b$,
$$ \max(a_1,\ldots,a_n) \leq b \quad \Leftrightarrow \quad a_1 \leq b \land \ldots \land a_n \leq b. $$
I.e. the maximum is less than some ceiling exactly if all the individual values are less than that ceiling.  So this explains the first step in the probability calculation at the end.
A: Yes, it is the random variable with the highest value.
For your second question, consider $n$ real numbers $x_1, \dots, x_n$. Suppose the largest of these is less than some value $x$. Then it follows that $x_1, \dots, x_n$ must all be less than $x$ as well. By independence
$$\mathbb{P}(M_n < x) = \mathbb{P}(X_1 < x, X_2 < x, \cdots , X_n < x) = \mathbb{P}(X_1 < x) \mathbb{P}(X_2 < x) \cdots \mathbb{P}(X_n < x)$$
and assuming that $X_1, \dots, X_n$ are all drawn from the same distribution (this was not stated in your question), it follows that
$$\mathbb{P}(X_1 < x) \mathbb{P}(X_2 < x) \cdots \mathbb{P}(X_n < x) = [\mathbb{P}(X_1 < x)]^n$$
because all of $\mathbb{P}(X_1 < x)$, $\mathbb{P}(X_2 < x)$, ..., $\mathbb{P}(X_n < x)$ would be identical.
A: I think it will help you to think through an example. Consider this one:
Let $x_1, x_2,$ and $x_3$ be three rolls with a fair die. Each $x_i$ can take the values from $1$ to $6$, however it is random which values they take.
Three Examples:
You roll $1$, $5$, and $2$ (i.e. $x_1 = 1$, $x_2 = 5$, $x_3 = 2$). Then $\max\{x_1,x_2,x_3\} = 5$.
You roll $1$, $1$, and $2$. Then $\max\{x_1,x_2,x_3\} = 2$.
You roll $3$, $2$, and $6$. Then $\max\{x_1,x_2,x_3\} = 6$.
Let us now calculate $\mathrm{Pr}(M_3<x)$ for, say, $x=4$. This means that all rolls must be below 4, i.e. 3 or lower. Each roll has $3/6 = 1/2$ chance of rolling $3$ or lower and the rolls are independent, hence the probability is given by
$$\mathrm{Pr}(M_3<4) = \frac12\times\frac12\times\frac12 = \frac18.$$
In other words
$$\mathrm{Pr}(M_3<4) = \left(\mathrm{Pr}(x_i<4)\right)^3. $$
If this example helps you understand it, it may help you generalize to other random variables.
