Does this limit converge on e? Playing around with some math in python today I came across what appears to be an interesting pattern:
Starting at n=1 as n approaches positive infinity, take (n+1)^(n+2)/n^(n+1) and get a list of ratios of exponential expressions. At first glance the ratios between the numbers appeared to be converging to something so...
Next, I took the difference between consecutive ratios, e.g. (n+2)^(n+3)/(n+1)^(n+2)-(n+1)^(n+2)/n^(n+1).
The differences appear to be approaching e (2.718...) as n gets larger.
The first few ratios rounded to the third decimal place are...
2^3/1^2 = 8
3^4/2^3 = 10.125
4^5/3^4 = 12.642
5^6/4^5 = 15.259
6^7/5^6 = 17.916
...
With their differences being...
10.125 - 8 = 2.125
12.642 - 10.125 = 2.517
15.259 - 12.642 = 2.617
17.916 - 15.259 = 2.657
...  
After the 13th iteration you get 2.711, and it looks like the series will converge on e as it gets arbitrarily large. That or likely positive infinity and my hunch is off!
Can anyone with better knowledge of limits tell me if I've stumbled across a novel way of calculating e or not?
Here's the python code for those curious (first loop stops at 15 because that's all my cheap phone could handle):

import numpy as np

ratios = []
i = 1
while i < 15:
    a = np.power(i,i+1)
    b = np.power(i+1,i+2)
    print(a)
    ratios.append(b/a)
    i+=1
print(ratios)

x=0
diffs = []
while x < len(ratios) - 1:
    temp = ratios[x+1] - ratios[x]
    diffs.append(temp)
    x+=1

print(diffs)


This reminds of the time I thought I discovered a novel formula for exponents about the golden ratio. I didn't, it was already known and I think this may be the case too but my brief search hasn't turned up anything yet.
Thanks!

We have
\begin{align*}
&\left( {1 + \frac{1}{{n + 1}}} \right)^{n + 2} (n + 2) - \left( {1 + \frac{1}{n}} \right)^{n + 1} (n + 1) \\ & = \left( {1 + \frac{1}{n}} \right)^{n + 1} \left[ {\left( {\frac{{(n + 2)n}}{{(n + 1)^2 }}} \right)^{n + 1} \frac{{(n + 2)^2 }}{{(n + 1)^2 }} - 1} \right](n + 1)
\\ &= 
\left( {1 + \frac{1}{n}} \right)^{n + 1} \left[ {\left( {1 - \frac{1}{n} + \frac{3}{{2n^2 }} + \mathcal{O}\!\left( {\frac{1}{{n^3 }}} \right)} \right)\frac{{(n + 2)^2 }}{{(n + 1)^2 }} - 1} \right](n + 1)
\\ &= \left( {1 + \frac{1}{n}} \right)^{n + 1} \left( {\frac{1}{n} - \frac{3}{{2n^2 }} + \mathcal{O}\!\left( {\frac{1}{{n^3 }}} \right)} \right)(n + 1) \\ &= \left( {1 + \frac{1}{n}} \right)^{n + 1} \left( {1 - \frac{1}{{2n}} + \mathcal{O}\!\left( {\frac{1}{{n^2 }}} \right)} \right) = e + \mathcal{O}\!\left( {\frac{1}{{n^2 }}} \right).
\end{align*}
