What is the value of $1^x$? I am trying to understand why $1^{x}=1$ for any $x\in\mathbb{R}$
Is it OK to write $1^{x}$? As the base 1 should not equal 1 for $1^{x}$ to be an exponential function? 
Is $1^{x}=1$ just because it is defined to be so?
If possible please refer me to a book or article that discusses this topic.

You don't have to specifically define $1^x$ to be something. It follows from the definition of exponentiation in general.
The story is roughly as follows: first there was exponentiation with exponents being natural numbers. That is just repeated multiplication of the base with itself. Then people noticed that this exponentiation obeys the rule
\begin{eqnarray}
a^{n+m} = a^na^m.
\end{eqnarray} This suggests that if we want to define $a^0$, then we want it to satisfy $a^m = a^{m+0} = a^ma^0$, so that it is natural to define $a^0 = 1$. Now, we can use the same rule to extend this operation to all integral exponents, i.e. positive and negative. Since we want $a^na^{-n} = a^{n-n} = a^0 = 1$, we are forced to define $a^{-n} = \frac{1}{a^n}$.
So, this way you have defined exponentiation with arbitrary integral exponents. Now, going back to the original exponentiation with natural numbers, you also notice that
\[(a^n)^m = a^{nm}
\] and $a^1 = a$. Applying the same considerations as above, you are forced to define $a^{\frac{1}{n}} = \sqrt[n]{a}$. This way, you have now extended exponentiation to all rational exponents. Note that all these rules imply that $1^x = 1$ for $x\in \mathbb{Q}$. Now, you extend all this to all real exponents using continuity, so you still have $1^x = 1\;\forall x\in \mathbb{R}$.
