# Proof: By Induction

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 $$n\ge 0$$.

### Special case $$n=0$$

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: $\exp_a(0) =\exp(0\cdot \ln(a))=\exp(0)=1=a^0.$

### Base case $$n=1$$

For $$n=1$$, we get (because the logarithm is the inverse function of the exponential function)

$\exp_a(1) =\exp(1\cdot \ln(x))=a=a^1.$

### Induction step $$n\to n+1$$

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

$\begin{array}{rcll} \exp_a(n+1)&=&\exp_a(1)\cdot \exp_a(n)&\text{by functional equation}\\ &=&a\cdot a^n&\text{by base case}\\ &=&a^{n+1}&\text{by definition of the n-th power function}.\end{array}$

Github: ### References

#### Bibliography

1. Forster Otto: "Analysis 1, Differential- und Integralrechnung einer Veränderlichen", Vieweg Studium, 1983