I know that if we have some independent random variables $x_1,x_2,x_3,\ldots,x_n$ having identical distributions:
$M_n=\max(x_1,x_2,x_3,\ldots)$ , then we say that probability that $M_n<x$ is $\Pr(M_n<x)$ then $\text{Pr}(M_n<x)$=$\text{Pr}(x_1<x,x_2<x,x_3<x,...)$=$(Pr(x))^{n}$
My question is, why does this not apply?
$\text{Pr}(M_n>x)$=${Pr}(x_1>x,x_2>x,x_3>x,...)$=$(1-Pr(x))^n$
As stated by Nap D.Lover, the issue is not related to probabilities, but to the definition of maximum.
With $M_n=\max\{x_1,\ldots,x_n\}$, we can say that $$\text{Pr}(M_n<x)=\text{Pr}(x_1<x,x_2<x,\ldots, x_n<x)=(Pr(x))^{n}$$
Because $M_n$ is smaller than $x$ if and only if each $x_i$ is smaller than $x$. Therefore the event $\{M_n<x\}$ is the same than the event $\{x_1<x\text{ and }x_2<x\text{ and }\ldots\text{ and }x_n<x\}$
$$\{M_n<x\}=\{x_1<x\text{ and }x_2<x\text{ and }\ldots\text{ and }x_n<x\}$$
But the fact $M_n$ is greater than $x$ only tells you that at least one $x_i$ is larger than $x$, not all of them. $$\{M_n>x\}=\{x_1>x\text{ or }x_2>x\text{ or }\ldots\text{ or }x_n>x\}$$
If you want to reverse the inequality you have to look at the minimum function instead of the maximum : let $m_n=\min\{x_1,\ldots,x_n\}$, then the event $\{m_n>x\}$ is the same than $\{x_1>x\text{ and }x_2>x\text{ and }\ldots\text{ and }x_n>x\}$, and therefore
$$\text{Pr}(m_n>x)=\text{Pr}(x_1>x,x_2>x,\ldots, x_n>x)=(1-Pr(x))^{n}$$