Proving the irrationality of $e^n$.

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 $e^3$.

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Wikipedia gives a sketch of a proof that $e$ is transcendental –  Henry Aug 27 '13 at 1:04
Edited my question. I don't want to use the fact that $e$ is transcendental, just as we don't need to in order to show that $e$ is irrational. –  nayrb Aug 27 '13 at 1:06
And as another note, I'm motivated by this question: math.stackexchange.com/questions/476899/…. –  nayrb Aug 27 '13 at 1:12
There is a proof for $e^r$ with $r\in\mathbb{Q}$ in Proofs From the Book by Martin Aigner and Günter M. Ziegler. You may want to look it up. –  EuYu Aug 27 '13 at 1:16
There are some nice notes by Keith Conrad in which in particular it is shown that if $r$ is a non-zero rational then $e^r$ is irrational (that's equivalent to your question). The proof is "elementary" but challenging. –  André Nicolas Aug 27 '13 at 1:17

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 $h \in \mathbb{Z}^+.$

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