Show $n^{1/n}$ is decreasing. This is getting to be a habit! I have done a fair amount of looking around however I cannot comment on other peoples questions yet in order to get further help when I don't understand something. My question is essentially this one Show $n^{\frac{1}{n}}$ is decreasing for $n \ge 3$. 
The answer chosen as the accepted solution helps me see where to begin, but I am unable to see where to go from there. Another thing is that in my question I am asked to show the sequence is decreasing for $n\ge3$ starting with $$(1+\frac{1}{n})^n \le n$$ for all $n\ge3$. So although the linked question provides some help, it doesn't work from this point which is why I'm not sure how to proceed.
I really appreciate the help!
Just want to say a massive thank you to Jorge, ab123 and Luiz for all your help, I don't even want to imagine how frustrating that must have been for you!
 A: We want to prove $n^{1/n} \geq (n+1)^{1/(n+1)}$ for $n\geq 3$.
This is equivalent to $n^{n+1}\geq (n+1)^n$
Which is equivalent to $(\frac{n+1}{n})^n\leq n$
Notice $(\frac{n+1}{n})^n=(1+\frac{1}{n})^n$.
So we have to prove $(1+\frac{1}{n})^n\leq n$ for $n\geq 3$.
Notice that $(1+\frac{1}{n})^n=\sum\limits_{i=0}^n \binom{n}{i}\frac{1}{n}^i$ by newton's theorem.
$\sum\limits_{i=0}^n \binom{n}{i}\frac{1}{n}^i=\sum\limits_{i=0}^n \frac{n!}{i!(n-i)!n^i}=\sum\limits_{i=0}^n \frac{n(n-1)\dots(n-i+1)}{i!n^i}\leq \sum\limits_{i=0}^n \frac{1}{i!}\leq 1+\sum\limits_{i=1}^n \frac{1}{2^{i-1}}\leq1+2\leq n$
A: I am not sure whether you are looking for a solution containing more powerful tools like derivatives. I try to show you this approach anyway. 
Consider the function $f(x) = x^{\frac{1}{x}}, x > 0$. When $x$ is a natural number this gives exactly your sequence.
Take $\log$ of both sides: $\log(f(x)) = \log(x^{\frac{1}{x}}) = \frac{1}{x}\log(x)$ and differentiate them:
$$\frac{f'(x)}{f(x)} = -\frac{1}{x^2}\log(x)+\frac{1}{x^2}$$
hence 
$$f'(x) = x^{\frac{1}{x}-2}(1-\log(x)).$$
Since $x^{\frac{1}{x}-2} > 0$, this is negative when 
$$1-\log(x) < 0 \rightarrow x > e.$$
So $f$ is decreasing when $x > e$. If you restrict to natural numbers, then $n^{\frac{1}{n}}$ is decreasing for $n \geq 3$.
A: Let $x=y^x,$ for $x\ge 1.$ Then we can show that $$\dfrac{dy}{dx}=\dfrac{y}{x^2}(1-\ln x).$$ It is not difficult to see that this implicit function gives a maximum for $y$ at $x=e$ then decreasing so that $y=1$ is a horizontal asymptote for the graph.
A: In this case, we have to show that a sequence $ (x_n)_{n \in \mathbb N} $ given by $ x_n := n^{1/n} $ is decreasing for $ n≥3 $. To do this, we should check that
$x_{n+1} \leq x_n $ this is $ (n+1)^{1/(n+1)} \leq n^{1/n} $.
raise both members of this inequality by $ n(n+1)$ we have
$ (n+1)^n \leq n^{(n+1)}$ 
$ \Leftrightarrow $ 
$ {(n+1)^n}/{n^n} \leq n $ 
$ \Leftrightarrow $ 
$ (1 + {1/n})^n \leq n $.
if you already know how to prove this last inequality you can use these equivalences to prove that the sequence is decreasing. 
another way of proving that $ \{(1 + {1/n})^n\}_{n \in \mathbb N} $ is decreasing and converges to the number $ e $:
the inequality $ {(b^{n+1} - a^{n+1})}/{(b-a)} < (n+1)b^{(n+1)} $ for all $ 0 \leq a < b $. 
To prove this use the binomial expansion
$ b^{n+1} - a^{n+1} = (b-a)(b^n +ab^{(n-1)} + a^2b^{(n-2)} + ... + a^{(n-1)}b +a^n) $
Then
$ {(b^{n+1} - a^{n+1})}/{(b-a)} < b^n +bb^{(n-1)} + b^2b^{(n-2)} + ... + b^{(n-1)}b +b^n = (n+1)b^{(n+1)} $.
Now with this inequality you can get
$ b^n [b - (n+1)(b-a)] < a^{n+1}$
if we set $a = 1 + 1/(n+1)$ and $ b = 1 + 1/n $, we have $ 0 \leq a < b $ and the term in brackets reduces to $1$ and we have
$ (1 + {1/n})^n < (1 + {1/(n+1)})^{n+1} $
so we prove that the sequence is decreasing. to show that the sequence converges, it suffices to show that it is bounded. We will again use the inequality above.
Set $ a =  1 $ and $ b = 1 + 1/(2n)$. This time the term in the brackets reduces to $1/2$, and we have
$ (1 + {1/(2n)})^n < 2 $
thus
 $ (1 + {1/(2n)})^{2n} < 4 $
but $ \{(1 + {1/n})^n\}_{n \in \mathbb N} $ is decreasing, then
$ (1 + {1/n})^n < (1 + {1/(2n)})^{2n} < 4 $ for all $n \in \mathbb N $.
Therefore the sequence $ \{(1 + {1/n})^n\}_{n \in \mathbb N} $ is decreasing and bounded. With this we conclude that $ \{(1 + {1/n})^n\}_{n \in \mathbb N} $ is is a convergent sequence and its limit is denoted by $ e $.
This answer complements the answer given by Jorge Fernández.
A: $n^{1/n} = \exp((1/n)\log(n)).$
Since $\exp(x)$ is strictly increasing it suffices to show that 
for $m >n$: 
$(1/n)\log(n) >(1/m) \log(m).$
$f(x) := (1/x)\log(x), x \ge 3.$
$f'(x) = -(1/x^2)\log(x) +1/x^2 .$
$f'(x) = (1/x^2)[1-\log(x)].$
$f'(x) <0$ for $x >3. $
$f(x) $ strictly decreasing for $x>3, $
$\rightarrow$:
For $m > n:$  
$n^{1/n} > m^{1/m}$, I.e. strictly decreasing.
