关于n趋于无穷时$(n + 1)^n$在$n$进制下的极限行为、末尾数字模式证明及相关文献的技术问询

Hey folks, I’ve been poking around a really interesting pattern with the polynomial expansion of $(n+1)^n$ when written in base $n$, and I’ve got a few questions I’m hoping the community can help unpack.

First, let’s lay out the observations I’ve made so far for finite $n$:

• When expressed in base $n$, $(n+1)^n$ consistently has a leftmost digit of 2, rightmost two digits of 01, with the middle coefficients varying.
• We can rewrite the expression using the limit definition of Euler's number:

  (n + 1)^n = n^n \left(1 + \frac{1}{n}\right)^n
  

• In base $n$, multiplying by $n^n$ essentially shifts the digits of $\left(1 + \frac{1}{n}\right)^n$ $n$ places to the right, keeping the underlying digit sequence intact.
• For all $n > 2$, $(n+1)^n$ has exactly $n+1$ digits (coefficients) when represented in base $n$.
• As $n$ grows larger, $\left(1 + \frac{1}{n}\right)^n$ approaches Euler's number $e$.
• The leading 2 and trailing 01 pattern comes from the symmetry of the binomial theorem—specifically the first term $1_n$ and last term $10_n$, plus carries from the "$1$" in $10_n$.

Core Questions I’m Curious About #

1. Limit Behavior as $n \to \infty$:
   How can we interpret the $2\ldots01_n$ pattern in relation to $e$ as $n$ approaches infinity? Is there a meaningful way to discuss the concept of $e$ in an "infinite base" system, or does this idea become purely abstract/theoretical?

2. Trailing Digit Pattern Proof:
   I’ve noticed another pattern in the last three digits:

   • For odd $n$, the trailing three digits are 101
   • For even $n$, the trailing three digits are $m01$ where $m = \frac{n}{2} + 1$
     Can we formally prove that this holds for all relevant $n$?

3. Relevant Literature:
   My searches haven’t turned up much research in this area—are there any papers, books, or established mathematical topics that delve into base-dependent polynomial expansion behaviors like this?

Mathematica Code for Exploration #

If you want to experiment with this pattern yourself, here’s the code to generate the digit sequences for $n$ from 2 to 100:

Table[{n, IntegerDigits[(n + 1)^n, n]}, {n, 2, 100}]

备注：内容来源于stack exchange，提问作者NeonNarwhal