Why can't a polynomial fit infinitely many points of an exponential function? 
Possible Duplicate:
Polynomial satisfying $p(x)=3^{x}$ for $ x \in \mathbb{N}$ 

I'm looking for an elementary solution to this question: 

There is no polynomial $P$ such that $P(0)=1, P(1)=3, P(2)=9, P(3)=27, \dots$.

 A: If $P(x)$ is a polynomial function of degree $d>0$ of $x$, then the differences $(\Delta P)(x)=P(x+1)-P(x)$ give a polynomial function of degree $d-1$ of $x$ (this follows for polynomials of one term, by the binomial formula, an the general case follows easily from this). You want a nonzero polynomial that satisfies $\Delta P=2P$, but the left hand side has degree one less than the righ hand side, so this is impossible.
This argument (as does the answer by marlu) does use the fact that a polynomial function, even restricted to its values at natural numbers, determines the underlying polynomial uniquely; otherwise the notion of degree (or leading term) of such a function would not be well defined. Note that over finite fields exponential functions are polynomial functions, in many differnt ways. I think it would be rather harder to give an argument not using the mentioned fact.
A: Assume $P(X) = a_n X^n + a_{n-1}X^{n-1} +\ldots + a_1 X + a_0$ is a polynomial that satisfies $P(k) = 3^k$ for all natural numbers $k$. Then $P(k+1) - 3P(k) = 0$ for all $k$, so the polynomials $P(X+1)$ and $3P(X)$ are equal. Now compare the highest coefficient.
