Let $X$ be the random variable obtained as the maximum of a throw of $m$ dice (each of which is $n$-sided). In other words, $X = \max\{l\_1,\cdots, l\_m\}$ where $l\_i$ can take any value between $1$ and $n$. Using the relationship between the sum of consecutive powers and Bernoulli numbers, one obtains the following formula for the probability of $X=k$:


$$Pr(X=k) = \frac{1}{n^m}[k^m -\frac{1}{m}B\_m(k) + \frac{1}{m}B\_m)]$$


where $B\_m$ is the $m$th Bernoulli number and $B\_m(k)$ is the $m$th Bernoulli polynomial evaluated at $k$. Clearly, if we let $n$ approach infinity, then the quantity above will approach zero. However, I am unsure what happens if we instead let $m$ approach infinity. The question then is does the following limit exist


$$Pr\_{\infty}(X=k) := \lim\_{m\to\infty}(\frac{1}{n^m}[k^m -\frac{1}{m}B\_m(k) + \frac{1}{m}B\_m)]) $$



Probability of the maximum of a throw of an infinite number of $n$-sided dice being $k$
Intuition should say that if we keep tossing a die with a finite number of outcomes, then sooner or later we will see every outcome, in particular, we will see the maximum outcome. A proof is


$$\Pr[\text{no throw equals $n$}] = \Pr[l\_1\neq n \text{ and } \dots \text{ and }l\_m\neq n]$$
 $$ = \Pi\_{j=1}^m \left(1-\frac{1}{n}\right) $$
 $$ = \left(1-\frac{1}{n}\right)^m $$
 $$ \to 0 \text{ as } m \to \infty .$$


So with probability approaching $1$, some throw equals $n$ and the maximum equals $n$ (so the probability that the max is $k < n$ goes to zero). I guess this doesn't say that the limit of your expression exists, but if it does, it's zero for $k < n$ and one for $k=n$.


(Edit: If you are interested in getting more comfortable with arguing this sort of thing, I like Mitzenmacher and Upfal's *Probability and Computing*, which is mostly about probability tools with applications to computing -- but then I'm in CS. Anyway, for finite cases, this sort of problem might be called "[balls into bins](http://en.wikipedia.org/wiki/Balls_into_bins)" or "[coupon collectors](http://en.wikipedia.org/wiki/Coupon_collector's_problem)".)


