(related to Proposition: Exponential Function of General Base With Natural Exponents)
We will prove by induction that the exponential function of general base \(a > 0\) is identical with the n-th power function, formally \(\exp_a(n)=a^n\) for all positive real numbers \(a > 0\) and all natural numbers \(n\ge 0\).
Using the corresponding proposition due to which \(\exp(0)=1\), we get from the definition of the exponential function of general base, and the definition of the n-th power function: \[\exp_a(0) =\exp(0\cdot \ln(a))=\exp(0)=1=a^0.\]
For \(n=1\), we get (because the logarithm is the inverse function of the exponential function)
\[\exp_a(1) =\exp(1\cdot \ln(x))=a=a^1.\]
Assume \(\exp_a(m)=a^m\) is correct for all \(m\le n\) for a given \(n\ge 0\). By virtue of the functional equation of the exponential function of general base, we get
\[\begin{array}{rcll} \exp_a(n+1)&=&\exp_a(1)\cdot \exp_a(n)&\text{by functional equation}\\ &=&a\cdot a^n&\text{by base case}\\ &=&a^{n+1}&\text{by definition of the n-th power function}.\end{array}\]