Let $n$ be a positive integer. I know the traditional proof that $e$ is irrational. How do we show that $e^n$ is irrational in some sort of similar line? I am of course assuming it is but I would be astounded if not. It occurs to me that since $e$ is transcendental, of course $e^n$ is irrational, but I don't want to use that fact.
Googling gives me something for $e^2$, but I could not easily find anything for $e^3$.
Niven's polynomials
Let $f:[0,1]\longrightarrow \mathbb{R}$ , $\displaystyle f(x)=\frac{x^n(1-x)^n}{n!}$ then
$$f(x)=f(1-x)$$
$$\displaystyle 0\le f(x)<\frac{1}{n!}$$
$$f^{(j)}(0)\;,\;f^{(j)}(1) \in \mathbb{Z} \;,\; j\ge 0$$
Proposition. The number $e^3$ is irrational.
Proof: Suppose that $\displaystyle e^3=\frac{a}{b}$
$$\displaystyle F=3^{2n}f-3^{2n-1}f'+3^{2n-2}f''-\cdots + f^{(2n)}$$
$$\displaystyle F'+3F=3^{2n+1}f$$
$$\displaystyle \mathbb{Z^+} \ni aF(1)-bF(0)=b\Bigl[e^{3x}F(x)\Bigr]_0^1=b\int_0^1 3^{2n+1}e^{3x} f(x)dx \longrightarrow 0^+\;,\;n \longrightarrow\infty $$ Contradiction , analogously $e^h$ is irrational for $h \in \mathbb{Z}^+.$