Prove that $(n+1)^{n-1}How would one prove that  $$(n+1)^{n-1}<n^n \ \forall n>1$$
I have tried several methods such as induction.
 A: Divide both sides of $(n+1)^{n-1}<n^n$ by $n^{n-1}$. We have now $$\frac{(1+1/n)^n}{1+1/n}<n.$$ The left side tends to $e$ from below. $e$ is less than 3 and the inequality holds for $n=2$.
A: For $n > 1$, apply GM $\le$ AM to following $n$ numbers $\big(\;\underbrace{n+1,n+1,\ldots,n+1}_{n-1 \text{ copies}}, 1\;\big)$. We have
$$((n+1)^{n-1}\cdot 1)^{1/n} \le \frac{(n-1)(n+1)+1}{n} = n
\quad\implies\quad  (n+1)^{n-1} \le n^n$$
Since the $n$ numbers are not the same, the inequality is strict and we are done.
A: For $n\gt1$, Bernoulli's Inequality gives
$$
\begin{align}
\frac{n^n}{(n+1)^{n-1}}
&=n\left(1-\frac1{n+1}\right)^{n-1}\\
&\ge n\left(1-\frac{n-1}{n+1}\right)\\
&=\frac{2n}{n+1}\\[6pt]
&\gt1
\end{align}
$$

We can also use Bernoulli's Inequality slightly differently. If $n\gt1$, then Bernoulli's Inequality is strict since $\frac1{n+1}\ne0$:
$$
\begin{align}
\frac{n^n}{(n+1)^{n-1}}
&=(n+1)\left(1-\frac1{n+1}\right)^n\\
&\gt(n+1)\left(1-\frac{n}{n+1}\right)\\[6pt]
&=1
\end{align}
$$
A: Let's try induction. Let n = 2. Then $(3)^{1}< 2^{2}=4$, which is clearly true. Now assume the statement is true for some natural number k where k >2. Now consider the case k+1. 
$((k+1)+1)^{(k+1)-1}<(k+1)^{k+1}$ = $(k+2)^{k}<(k+1)^{k+1}$ 
Let $k= l-1$ where $l$ is some natural number larger then k. Then: 
$(k+2)^{k}<(k+1)^{k+1}$ = $(l+1)^{l-1}<l^{l}$
Which is what we wanted to show. So now the inductive step gives the desired result. 
(Why don't I feel confident of this proof even though I just wrote it? Still,that's my best crack at it.) 
A: Proof By Induction: The hypothesis is true for $n=2$. Let it be true for $n=k$. Then  $$(k+1)^{k-1}<k^k\\\implies \left(1+\frac{1}{k}\right)^{k}<{k+1}$$Now, $$\frac{(k+2)^k}{(k+1)^{k+1}}=\left(1+\frac{1}{k+1}\right)^{k}\frac{1}{k+1}<\left(1+\frac{1}{k}\right)^{k}\frac{1}{k+1}<1$$
Proof By Calculus: Let us define the function $f:\mathbb{R}^+\to \mathbb{R},\ f(x)=(x-1)\ln (x+1)-x\ln x$. Then, since $f(1)=0$, we need to show that $f$ is decreasing on $[1,\infty)$. Note that $f'(x)=\ln (x+1)+\frac{x-1}{x+1}-\ln x-1=\ln (x+1)-\ln x-2/(x+1)\implies f''(x)=\frac{1}{x+1}-\frac{1}{x}+\frac{4}{(x+1)^2}=\frac{x(x+1)-(x+1)^2+4x}{x(x+1)^2}=\frac{3x-1}{x(x+1)}>0\ \forall x\in [1,\infty)\implies f'(x)$ is increasing in $[1,\infty)\implies f'(x)<\lim_{x\to \infty}f'(x)=0\ \forall x\in [1,\infty)\implies f$ is decreasing in $[1,\infty)$.
A: My try to prove $\forall n>1 : (n+1)^{n-1} < n^n$:
$$(n+1)^{n-1} < n^n \Leftrightarrow (n+1)^{n} < n^n (n+1) \Leftrightarrow \left(\frac{n+1}{n}\right)^n < n+1$$
Now apply induction on this result:
Base case: $n = 2 \Rightarrow \left(\frac{3}{2}\right)^2 < 3 \Rightarrow 2.25 < 3$ OK
Induction hypothesis: assume $\left(\frac{n+1}{n}\right)^n < n+1$ true for all $n \leq k$
Induction step: we prove that $\left(\frac{n+1}{n}\right)^n < n+1$ is true for $n = k+1$
$$\left(\frac{k+2}{k+1}\right)^{k+1} < k+2 \Leftrightarrow \left(\frac{k+2}{k+1}\right)^k < k+1$$
Now we have that $\left(\frac{k+1}{k}\right)^k > \left(\frac{k+2}{k+1}\right)^k$ and because of the induction hypothesis, we have proven by induction that $\forall n>1 : \left(\frac{n+1}{n}\right)^n < n+1$ and thus that $\forall n>1 : (n+1)^{n-1} < n^n$
A: For completeness' sake, here is a proof coming from combinatorics.  The terminology and original proof can be attributed to Konheim and Weiss (1966) and the proof reproduced here by Pollak.  I first encountered the problem in Stanley's textbook Enumerative Combinatorics.  See also here for more information and references.
Imagine the scenario of a parking lot with $n$ labeled spaces all in a row where traffic may only flow in one direction through the parking lot.  You have $n$ cars, each of whom have their preferred space in the lot.  The sequence of preferences we call a preference sequence.  When a car enters the lot, they drive until they have reached their preferred space.  If it is unoccupied, they park in the space.  Otherwise, they continue driving and park in the next available unoccupied space.  If all spaces from that point on were occupied, then they leave and were unable to successfully park.
It is clear that some preference sequences lead to all cars successfully parking (for trivial example, when everyone prefers parking in the first space by the entrance) and is also clear that some preference sequences lead to few cars successfully parking (for trivial example, when everyone prefers parking in the last space by the exit).  We call a preference sequence a parking lot sequence or parking lot function if the preference sequence leads to all cars successfully parking.
We ask the question of how many parking lot sequences there are for $n$ cars and $n$ spaces.
To see this, imagine briefly that you were to wrap the row of parking spaces around to form a circle and that you were to add an additional $n+1$'st space.  The cars act the same as otherwise however now rather than exiting once passing the $n$'th space they continue and loop around the circle taking one of the otherwise earlier spaces.  It is clear that every preference sequence here leads to all cars parking in this circular lot and that exactly one space will be left unoccupied.  Further, it is clear that a preference sequence for this circular lot will be a parking lot sequence if and only if that unoccupied space is the newly added $n+1$'st space.  Finally, by symmetry, there will be exactly as many preference sequences which leave one space unoccupied at the end as any other.
It follows then that the number of parking lot sequences is $(n+1)^n / (n+1) = (n+1)^{n-1}$.
The probability of a preference sequence being a parking lot sequence for our original lot then, if the preference sequences are each equiprobable, will be:
$$\frac{(n+1)^{n-1}}{n^n}$$
Since this is a probability, we know it to be between $0$ and $1$.  Since for $n>1$ it is possible for a preference sequence to not be a parking lot sequence we know the probability can not be exactly $1$.
This proves then that $(n+1)^{n-1}<n^n$ for all $n>1$
