Why do mathematicians need to define a zero exponent? I think a zero exponent can be defined logically from various premises; 


*

*We can define an exponent as the number of times the base "appears" in a multiplication process.
Thus: $2^3 = 2*2*2 $,  $2^1 = 2 $, $2^0 = (   )$ doesn't appear at all.  


Here's the thing, I know that we can literally define $2^{(something)}$ to be anything and as long as we are consistent with our definition, we can show that any result follows directly from our rules and definitions. 


*But mathematicians observed (and defined) a very nice property of exponents, namely; $b^n * b^m = b^{n+m}$
Thus, if we we want to be consistent with this definition AND WE WANTED TO INCLUDE THE NUMBER ZERO IN THIS, and we put n=0, then we have that $b^0 * b^m = b^m$
And since we have already defined this process 
              $Anything * 1 =  anything   .....(*)$
Then it should follow that in order to be consistent with our rules, bexp 0 has to be defined as 1.
(ofcourse if   (*) was defined such as $anything * 5 = anything$, then we'd have to say that $b^0 = 5$). 


My question is, however, why do we need to check the case when $n=0$? Why is it important/helpful? What problems will we face if we simply ignored $0$? I mean, the number $0$ wasn't even defined before? ofcourse the number $0$ alone is very useful in other contexts, but why here? 
Lastly, fractional exponents are defined in such a way that they "act" like roots; thus making writing and dealing with roots easier, negative exponents make quotients easier and so on.. Is there some situation where we would for example get (something raised to the zero) as a result and therefore we would have to KNOW what that must mean?
I'm writing from a mobile version of the website so I can't use the Math symbols. Sorry about that.
Thank you. 

In a world where we define exponentiation (of positive) by zero:

Compute $$ \int_a^b e^x \, \mathrm{d}x $$

An antiderivative of $e^x$ is $e^x$, so $\int_a^b e^x \, \mathrm{d}x = e^b - e^a$

In a world where we don't

Compute $$ \int_a^b e^x \, \mathrm{d}x $$

We prove by cases that the answer is
$$\int_a^b e^x \, \mathrm{d}x = \begin{cases}
e^b - e^a & a \neq 0, b \neq 0
\\ 1 - e^a & b = 0, a \neq 0
\\ e^b - 1 & b \neq 0, a = 0
\end{cases} $$
Case 1: $0 < a \leq b$ or $a \leq b < 0$
An antiderivative of $e^x$ is $e^x$, so $\int_a^b e^x \, \mathrm{d}x = e^b - e^a$
Case 2: $0 = a < b$
This is an improper integral which we compute as
$$ \int_a^b e^x \, \mathrm{d}x = \lim_{t \to 0^{+}} \int_t^b e^x \, \mathrm{d}x = \lim_{t \to 0^{+}} (e^b - e^t) = e^b - 1 $$
Case 3: $a < b = 0$
This is an improper integral which we compute as
$$ \int_a^b e^x \, \mathrm{d}x = \lim_{t \to 0^{-}} \int_a^t e^x \, \mathrm{d}x = \lim_{t \to 0^{-}} (e^t - e^a) = 1 - e^a $$
Case 4: $a < 0 < b$
$$ \int_a^b e^x \, \mathrm{d}x
= \int_a^0 e^x \, \mathrm{d}x + \int_0^b e^x \, \mathrm{d}x
= (1 - e^a) + (e^b - 1) = e^b - e^a $$
Case 5: $a = b = 0$
$$\int_a^b e^x \, \mathrm{d}x = \int_0^0 e^x \, \mathrm{d}x = 0 $$
Case 6: $b < a$
$$\int_a^b e^x \, \mathrm{d}x
= - \int_b^a e^x \, \mathrm{d}x  $$
and look up the right hand side by appealing to the appropriate case to determine
$$\int_a^b e^x \, \mathrm{d}x = \begin{cases}
e^b - e^a & a \neq 0, b \neq 0
\\ 1 - e^a & b = 0, a \neq 0
\\ e^b - 1 & b \neq 0, a = 0
\end{cases} $$

