What is $ {e^0}? $

Hint: An exponential function is defined as a function with a positive constant other than $ 1 $ raised to a variable exponent. A function is evaluated by solving at a specific input value. An exponential model can be found when the growth rate and initial value are known. The power series expansion of the exponential function. Let's represent the exponential function $ f(x) = {e^x} $ by infinite polynomials. The exponential function is an infinitely differentiable function. First we expand the exponential function and put the given data to get the answer.

Complete step-by-step answer:
First we find the expansion of exponential function $ f(x) = {e^x} $
Therefore $ f(x) = {e^x} = 1 + x + \dfrac{{{x^2}}}{{2!}} + .... $
We have from the given data $ x = 0 $ ,put this in above expansion and get the result.
 $ \Rightarrow f(x) = 1 + 0 + \dfrac{0}{{2!}} + ... $
 $ \Rightarrow f(x) = 1 $
Therefore the value of $ {e^0} = 1 $
So, the correct answer is “1”.

Note: We have to solve this problem in an alternative style like, we have $ {e^0} $ . We know $ {a^x} \times {a^y} = {a^{x + y}} $ , then $ {e^0} = {e^{1 - 1}} $
 $ = {e^1} \times {e^{ - 1}} $
We know $ {e^{ - 1}} = \dfrac{1}{{{e^1}}} $ , use this in above line and we get
 $ = \dfrac{{{e^1}}}{{{e^1}}} $
 $ = 1 $
Therefore the value of $ {e^0} = 1 $ .
Formula of expansion of exponential function $ {e^x} $ is $ {e^x} = 1 + x + \dfrac{{{x^2}}}{{2!}} + .... $