Analyses with Powers and Transcendentals
Problem Introduction
Which is greater, $88^{89}$ or $89^{88}$? Computing with straightfoward arithmetic, we have that $88^{89} > 89^{88}$. But is there another way to look at the problem?
Notice that $89 = 88 + 1$. Substituting $88 + 1$ for $89$, we get $88^{88 + 1} > (88 + 1)^{88}$. To generalize the question further, we replace $88$ with a natural number $x$, asking ourselves “under what condition does $x^{x+1} > (x+1)^x$ hold?”
Quick Observation
Consider the set $A = \{0, 1, \dots, 9\}$. Computing the left and right expressions of the inequality for each $x \in A$, we have
$\mathbf{x}$  $\mathbf{x^{x+1}}$  $\mathbf{(x+1)^x}$ 

0  0  1 
1  1  2 
2  8  9 
3  81  64 
4  1,024  625 
5  15,625  7,776 
6  279,936  117,649 
7  5,764,801  2,097,152 
8  134,217,728  43,046,721 
9  3,486,784,401  1,000,000,000 
Observe that $\forall x \in A$ such that $x \leq 2$, $x^{x+1} < (x+1)^x$. Now observe that $\forall x \in A$ such that $x \geq 3$, $x^{x+1} > (x+1)^x$. Next, we perform similar observations for the set of all nonnegative real numbers.
Analysis with e
We prove that $x^{x+1} > (x+1)^x$ for any real number $x \geq 3$ by finding the stricter condition that $x^{x+1} > (x+1)^x$ for any real number $x > e$.
Suppose $x \in \mathbb{R_{> 0}}$. Dividing $x^x$ on both sides of the inequality, we get
Let $h : \mathbb{R_{> 0}} \rightarrow \mathbb{R_{> 0}}$ be given by $h(x) = (1 + \frac{1}{x})^x$. Note that $\lim_{x\to\infty} h(x) = e$. To prove that $e$ is the least upper bound of $h$, we show that $h$ always increases when $x > 0$. We use the following method borrowed from an online thread:^{1}
It is sufficient to show that $ln(h(x)) = ln(1 + \frac{1}{x})^x$ is increasing. Let $y = \frac{1}{x}$. Substituting in $y$, the expression can be rewritten as
$\begin{align} \frac{ln(1+y)  ln1}{y} \end{align}$Now we show that as $y$ decreases, (1) increases. Notice that $h$ is strictly positive when $x > 0$, so $ln(h(x))$ is concave for $x > 0$. Also notice that (1) is the slope of the line through $(1, ln1)$ and $(1+y, ln(1+y))$. Since any concave function has the property that the slope of the line (1) increases as $y$ decreases, $h$ always increases for $x > 0$.
Since $h$ converges to $e$ and always increases for $x > 0$, we have that $e$ is the least upper bound of $h$. Since $e$ is the least upper bound of $h$, then for any $x \in \mathbb{R}$ such that $x \geq e$, we have that $x \geq (1 + \frac{1}{x})^x$. Since this inequality is equivalent to $x^{x+1} > (x+1)^x$, we get
$\square$
Analysis with Newton’s Method
Let $f(x) = x^{x+1}$ and $g(x) = (x+1)^x$. Note that $x^{x+1}$ and $(x+1)^x$ are polynomials, so $f$ and $g$ are continuous. We prove a stronger result by finding the only point of intersection of $f$ and $g$ in the nonnegative reals.
Consider the interval (0, e). From the sections above, we know that the inequality between $x^{x+1}$ and $(x+1)^x$ flips when going from $x = 0$ to $x = e$. Since $f$ and $g$ are continuous, then with the Intermediate Value Theorem (IVT), we have that for some $c \in (0, e)$, $f(c) = g(c)$.
To find $c$, we evaluate the expression $f(x) = g(x)$, which is given by $x^{x+1} = (x+1)^x$. Unfortunately, there seems to be no straightforward way to apply the Lambert W function.^{2} Instead, we consider the interval Newton’s method, a modification of Newton’s method to find all intersections in a given interval.^{3} Applying this method on (0, e), we approximate a single intersection $c = 2.293166\dots$, which (as an aside) is curiously a transcendental number.^{4}
Now we show that $\forall x > c, x^{x+1} > (x+1)^x$:
Suppose $\exists x > c, x^{x+1} <= (x+1)^x$ to reach a contradiction.
Consider the case $x^{x+1} = (x+1)^x$. Since $x^{x+1} > (x+1)^x$ for $x > e$, then $x$ must be an intersection in the interval $(c, e)$. But from interval Newton’s method, we know there cannot be another intersection in $(c, e)$. $\Rightarrow\Leftarrow$
Consider the case $x^{x+1} < (x+1)^x$. Since $x^{x+1} > (x+1)^x$ for $x > e$, then $x$ must be an intersection in the interval $(c, e)$. From interval Newton’s method, we know there cannot be another intersection in $(c, e)$. But by IVT, there must be another intersection in $(c, e)$ since $f(x) < g(x)$ and $f(x') > g(x')$ for some $x' > e$. $\Rightarrow\Leftarrow$
So by contradiction,
Moreover, $x^{x+1} < (x+1)^x$ for $0 \leq x < c$ can be proved via similar “proof by cases” as above, which will be left as an exercise.
$\square$
Footnotes

zhw. Answer to “How to prove (1 + 1/x)^x is increasing when x > 0?” https://math.stackexchange.com/questions/83035/howtoprove11xxisincreasingwhenx0 ↩

Miquilino, Alexandre Ribeiro. Answer to “How do you solve x^(x+1) = (x+1)^x?” https://www.quora.com/Howdoyousolvexx1x1x ↩

Seidl, Daniel. “Formalisation of Interval Methods for Nonlinear RootFinding” https://www21.in.tum.de/students/past/interval_newton/assets/newton.pdf ↩

Lord, Nick. “86.16 Two Other Transcendental Numbers Obtained by (Mis)Calculating e.” The Mathematical Gazette, vol. 86, no. 505, 2002, pp. 103–05. https://doi.org/10.2307/3621590 ↩