Having problems with Exponents $1^n =1$ where n is a positive integer I understand.
But $1^{1/2}$ or $1^{0.5}$ is also $1$, I have difficulty in understanding that.
And does that mean $1^n$ where n is any positive real no equals $1$?
$2^{-5}$ = $1 / 2^5$ = $1/32 =0.03125 $
I can compute this but I really don't understand this law.How can you raise something to negative power and if you can how is it obtained by removing the $-$ sign and dividing it by $1$ ?                                          
 A: As JMoravitz said, defining $a^n$ as "a multiplied by itself n times" only makes sense for n a positive integer.  We are free to define $a^x$, for x not a positive integer as we please- and we would like to do that so that it has "nice" properties.
One of the "nice" properties that $a^n$ is that $a^{n+ m}= a^{n}a^m$.  To see that imagine writing n+ m copies of n in a row.  That can be separated as the first n copies times the last m copies: $(a^n)(a^m)$.
How should we define $a^0$? Obviously we can't multiply a by itself "0 times"!  But 0 is the additive identity.  For any n, n+ 0= n.  If we want $a^{n+m}= (a^n)(a^m)$ true even when m= 0 we must have $a^n= a^{n+ 0}= (a^n)(a^0)$.  In order to have $a^n= (a^n)(a^0)$ we must define $a^0= 1$.
"Negative integers" are the additive inverses of the positive integers.  Any negative integer can be written "-n" for some positive integer n such that n+ (-n)= 0.  Then we want $a^{n+(-n)}= (a^n)(a^{-n})= a^0= 1$.  So we must define $a^{-n}= \frac{1}{a^n}$. For that to make sense we must require that a is not 0.
The set of rational numbers is defined "multiplicatively": if r is any non-zero number then there exist $\frac{1}{r}$ such that $r\left(\frac{1}{r}\right)= 1= r^0$.  For that we need a multiplicative property.  For positive integers, m and n, and any number a, $(a^m)^n= a^{mn}$.  to see that, think of $(a^m)^n$ as n rows, each row consisting of m copies of a.  There is a total of mn a's in that "rectangle" of a's.
Now, we want to define $a^{1/n}$ so that property still holds.  That is, we want $\left(a^{1/n}\right)^n= a^{(1/n)n}= a^1= a$. In order that this be true we have to define $a^{1/n}= \sqrt[n]{n}$, the "principle" nth root which, as long as a is a positive real number, is a positive real number.  And, in order that this be "well defined", we must require that a be a positive real number.
Finally, we have to define $a^x$ for x any real number. The set of real numbers cannot be defined "algebraically"- they have to be defined "analytically", with some kind of limit process.  The simplest way to do that is to define a real number as an equivalence class of sequences of rational numbers.  For example, we can think of the real number $\pi$ as defined by the sequence of rational numbers {3, 3.1, 3.14, 3.141, 3.1415, 3.14159, …}.  For any positive a, we would define $a^{\pi}$ as the limit of the sequence $\{a^3, a^{3.1}, a^{3.14}, a^{3.141}, a^{3.1415}, a^{3.14159}, …\}$.  
A: It's a matter of definition and fitting things into the definition.
Consider this process of learning.
Definition 1:  If $n \in \mathbb Z$ and $n \ge 2$ then $b^n = \underbrace{b\cdot b\cdot..\cdot b}_{n\text{ times}}$.
From there it's easy to verify that $b^nb^m = b^{n+m}$ and that $(b^n)^m = b^{n*m}$ and if $n \ge m+2$ the $\frac{b^n}{b^m} = b^{n-m}$.  This becomes our fundamental law:
Fundamental law:  $b^nb^m = b^{n+m}$ and $(b^n)^m = b^{nm}$.
We want to extend our definition of $b^x$ from Definition 1: where we have it it defined for $x\in \mathbb Z$ and $n \ge 2$ to $x\in \mathbb R$, but where a) if $x \in \mathbb Z$ and $x \ge 2$ our new definition will give the same value as Def 1 and b)  Our fundamental law must be true.
For us to do this we must have:  $b^0*b^k = b^{k+0} = b^k$.  That means $b^0$ must equal $1$ and we must have $b^{-k}b^{k} = b^{-k+k} = b^0 = 1$.  This means that $b^{-k} = \frac 1{b^k}$.  (This also means that we can define $0^n$ for $n\in Z$.
So we have 
Definition 2:  If $b\ne 0$ then $b^0 = 1$.  $b^1 = b$ and if $k \in \mathbb Z; k \ge 2$, $b^k$ is defined as in Definition 1; but if $k \in \mathbb Z$ and $k < 0$ then $k = -|k|$ and $b^{k} = b^{-|k|} = \frac 1{b^{|k|}}$.
Now our fundamental law says that $(b^n)^m = b^{nm}$ so $(b^{\frac 1k})^k = b^{\frac 1k * k} = b^1=b$ so we must have that $b^{\frac 1k}$ be one of the $k$th roots of $b$.
Now only positive numbers are guaranteed to have roots so this leads to definition three:
Definition 3:  If $b > 0$ and $k \in \mathbb Q$ and $k = \frac mn$ where $m,n\in \mathbb Z$ then $b^k =  b^{\frac mn} = \sqrt[n]{b^m}$ where $m \in \mathbb Z$ and $b^m$ is defined as in Definition 2.
Now, I'm going to wave my hands in discussing $b^x$ where $x \in \mathbb R$.  Every real number $x$ will have a sequence of rational numbers $q_i$ where $q_i \to x$.  If it's easier to think about it, you can consider the $q_i$ as the decimal expansions $x$.  If $q_i\to x$ then (you'll have to trust me on this) $b^{q_i} \to M$ for some real number $M$.  
Definition 4:  If $b > 0$ and $x \in \mathbb R$ and $\lim\limits_{n\to \infty} q_n = x$ for some sequence of rational $q_n$ then $b^x$ will be defined as $\lim\limits_{n\to \infty}b^{q_n}$ where $b^{q_n}$ is defined as in definition 3.
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So... to your questions.
If $1^{\frac 12} = w$ then $(1^{\frac 12})^2 = 1^1 = 1 =w^2$ so $w =\sqrt 1 = 1$.
"And does that mean $1^n$ where n is any positive real no equals 1?"  
Yes.  It does.  And $n$ doesnt just have to be positive real. $n$ could be zero or negative.
And what about $2^{-5}$?
If $2^{-5} = w$ then then $2^5*2^{-5} = 2^5 * w$ and so $2^5*2^{-5}=2^{5-5}=2^0 = 1 = 2^5*w = 32w$.  So $w = 2^{-5} = \frac 1{32}$.
A: The general definition is given by $$a^x=e^{x\ln(a)}$$ so $$2^{-5}=e^{-5\ln(2)}$$
A: $$x^{-1} = \dfrac{1}{x}, x^{-n} = (x^{-1})^n, x^{m/n} = \sqrt[n]{x^m}$$ are all definitions. The reason for the notation is to work nicely with the laws of exponents you already know. The intuition for $2^{-5}$ no longer means "multiply -5 copies of 2 together".
A: As David Peterson stated, $x^{-n}=\frac{1}{x^n}$ is a definition. 
I want to give an easy motivation why one would chose $x^{-n}$ to be defined as $\frac{1}{x^n}$. 
First notice, that you want all the rules you know for exponents also to work with negative exponents. So $x^{-n}$ should be equal to $(x^n)^{-1}$. 
Take now the product $x^n\cdot x^{-n}$. If we want all the laws of exponents to work with the definition of $x^{-n}$, this product could be rewritten in the following way: $x^n\cdot x^{-n}=(x^n)^1\cdot (x^n)^{-1}=(x^n)^{1+(-1)}=(x^n)^0=1$. So we have $x^n\cdot x^{-n}=1$, which is why you choose $x^{-n}$ to be $\frac{1}{x^n}$.
A: $1^N = 1$ for any real value of N.
You are OK with the case of N being integer. This is good start.
The case where $N$ is not integer, as in your example: $1^{0.5}$ (also referred to as $\sqrt{1}$) could be justified as follows:
What is the number $x$ that if you multiply it by itself, you get 1?
This calls for solving the equation $$x^2=1$$ Taking the square root of both sides, we get two numbers that satisfy the equation, namely the number $+1$ and the number $-1$. 
That is for the $+1$ case: $$\sqrt{x^2}=\sqrt{1}=1^{0.5}=1$$
I hope that his is somewhat convincing.
As for the negative powers part, think of how you would do $\frac{4^3}{4^5}$, you may want to write it down as where the "." is a multiplication sign:
$$K=\frac{4^3}{4^5} = \frac{4.4.4}{4.4.4.4.4}=\frac{1}{4.4}$$
If you notice the powers of the numerator (3) and the denominator (5) and compare them with the power of the denominator (2), you may observe that $5-3=2$. That is we could write:
$$K=\frac{4^3}{4^5}=\frac{1}{4^{5-3}}=\frac{1}{4^{2}}$$
Suppose now we want to calculate $Z$ by multiplying $K$ by $4^7$:
$$Z=4^{7}\frac{1}{4^{2}}$$
Using the above approach:
$$Z=4^{7}\frac{1}{4^{2}}=\frac{4.4.4.4.4.4.4}{4.4}=4.4.4.4.4=4^5$$
Observe the powers here:
$$Z=4^{7}\frac{1}{4^{2}}=4^5$$
Do you see that $7-2=7+(-2)=5$?
The (-2) here is what you use the negative power for.
That is why we can do:
$$Z=4^{7}.4^{-2}=4^{7-2}=4^5$$
and we can do $$K=\frac{4^3}{4^5}=4^{3-5}=4^{-2}=\frac{1}{4^2}$$
This may be a good read: What do Fractional Exponents mean
