Prove that as more dice are rolled and summed, any particular outcome becomes vanishingly unlikely

Question

"Suppose $n$ dice are rolled and their sum recorded; let $P_m(n)$ be the probability that this sum is $m$. Let $P(n) = \displaystyle \max_m P_m(n)$. Prove that $P(n) \to 0$ as $n \to \infty$."

This is a question I came up with myself (although I doubt it's original), and figured there would be a readily conceivable, slick solution even to someone who's not well versed in probability/statistics. Alas after playing with generating functions and the ilk for a little, I'm left nowhere better than where I started... which is, "This is almost certainly true by the Central Limit Theorem." But again, almost no background in statistics, so I would like a more basic answer if possible.

Answer 1

Extracting/merging answers from this question and this one, there are two related approaches

1. The sum of discrete uniform random variables (i.i.d.) is asymptotically normal. Otherwise, the exact distribution is that of a normalized extended binomial coefficient, see "Polynomial Coefficients and Distribution of the Sum of Discrete Uniform Variables".
2. In general generating functions will give you the probabilities. For example for a 6-sided die, the coefficients in the expansion of $$\frac{x(1-x^6)}{6(1-x)} = \frac16 x^1 +\frac16 x^2 +\frac16 x^3 +\frac16 x^4 +\frac16 x^5 +\frac16 x^6$$ give the probabilities for each specific sum (where the sum is the exponent). You can then find the probabilities of sums when throwing several independent dice by multiplying their generating functions $$\left(\frac{x(1-x^6)}{6(1-x)}\right)^n=6^{-n}\left(x+x^2+x^3+x^4+x^5+x^6\right)^n$$

As explained in the document, these are called polynomial coefficients, or more often multinomial coefficients.

$$ (x_1 + x_2 + \cdots + x_m)^n = \sum_{k_1+k_2+\cdots+k_m=n} {n \choose k_1, k_2, \ldots, k_m} \prod_{t=1}^m x_t^{k_t}\,$$ where $${n \choose k_1, k_2, \ldots, k_m} = \frac{n!}{k_1!\, k_2! \cdots k_m!}$$

We can use this facts to deduce that $P(n)$ tends to $0$. However there might be a smarter/easier way to do so without computing polynomial coefficients.