BODMAS (exponentiation and brackets) I’m going to try and be as articulate as possible, so if anything needs to be clarified, please don’t hesitate to let me know.
If we have $(-5)^2$, we obviously get $25$. Now, according to the rules of BODMAS we should ideally solve the bracket and then deal with the exponential but in this case we deal with the power first by doing:
$(-5) \cdot (-5)$
And then we eliminate the brackets by doing $(-5 \cdot -5) = 25$.
My question is, am I missing something or are the rules of BODMAS loose or have exceptions as such? I searched the internet but could not find anything to answer my question.
Looking forward to the responses!
Thank you in advance.
 A: The interpretation is supposed to be unique. BODMAS is not the clearest way to explain it in this case. Here the confusion arises as the negative sign is part of the value rather than a change-of-sign operation:
$$
(-5)^2=-5\cdot(-5)=25\\
-5^2=-5\cdot 5 = -25
$$
The first expression is a single value, negative 5, which is put into a single operation, squaring. The second expression is the number 5 put into a squaring operation and the result put into a change-of-sign operation.
Hence the use of brackets AND of the minus symbol are to be interpreted differently here.
Brackets are often used, when a negative number occurs within other operations. As seen in the multiplication above, where the brackets around the last $-5$ is used to separate the multiplication dot and the non-operator-like sign of the value.

Also note that if you try evaluating the brackets in the first expression, then we have:
$$
(-5)=-5
$$
so the value that should be squared is still negative 5 if you try to look at it from this perspective.

BODMAS also sometimes proves less relevant when working on algebraic reduction. For instance:
$$
(x+2)^2=(x+2)\cdot(x+2)=etc.
$$
here you can make no further reduction of the group $x+2$ inside the brackets, and so you use this result as a group in the ongoing evalution.
AND one may even refuse to perform reductions before using brackets as a group:
$$
(x+x)^2=(x+x)\cdot(x+x)=\underbrace{x^2}_{F}+\underbrace{x^2}_{O}+\underbrace{x^2}_{I}+\underbrace{x^2}_{L}=4x^2
$$
whereas BODMAS would like you reduce the bracket first (as much as possible):
$$
(x+x)^2=(2x)^2=2x\cdot 2x=4x^2
$$
and luckily both ways to structure the evaluation agree on the final reduction!
