Prove $5^n > n^5 $ for $n≥6$ How can I prove
$5^n > n^5 $ for $n≥ 6  $
using mathematical induction ?
So my solution is
Step 1: For $n = 6$
$5^6 = 15625$, $6^5 = 7776$
then $5^n > n^5 $ is true for $n = 6  $
Step 2: Assume it's true for $n = k$,  $5^k > k^5 $
Step 3: Prove it's true for $n = k+1$
$5^{k+1} > (k+1)^5 $
Here when things get messy I don't know how to prove it for $n = k+1$
 A: It is sufficient to consider the function $\frac{\ln x}{x}$ for $x>0$ and discuss its maximum and monotonicity. The answer should be as follows.

*

*When $0<x<e$, the function $\frac{\ln x}{x}$ is increasing.

*When $x>e$, the function $\frac{\ln x}{x}$ is decreasing.

From these conclusions, the inequality $5^n>n^5$, which is equivalent to $\frac{\ln5}{5}>\frac{\ln n}{n}$, for $n\ge6$ is derived immediately.
A: $$P(n): 5^n > n^5   \text{  :   } \forall n \ge6$$

*

*Base Case: $5^6 >6^5$ is True


*Inductive Case:$P(k)\implies P(k+1)$
The induction step, proves that if the statement holds for any given case $n = k$, then it must also hold for the next case $n = k + 1$. These two steps establish that the statement holds for every natural number $n$. The base case does not necessarily begin with $n = 0$, but often with $n = 1$, and possibly with any fixed natural number $n = N$(here, $N = 6$), establishing the truth of the statement for all natural numbers $n\ge N.$
$$\color{blue}{P(k): 5^k > k^5 \implies  (5^k5 = 5^{k +1})>5k^5} $$
To prove $P(k+1) $ is true we need to prove $(5^{k+1} = \color{blue}{5^k.5  > 5k^5} )> (k+1)^5$
Now,
$\color{blue}{1< 5k < 10k^2 < 10k^3< 5k^4}$ is true for $k\ge3 \implies \forall k\ge6$ You can prove this by considering that $P(1<n<5)$ is not true but we know this is obvious.
$k^4 = k^4 \implies \color{red}{5k^4 = 5k^4} \text{ for } (k\ge 1 \implies \text{ also true for } k\ge 6)$
Consider  $\color{red}{5k^4 = 5k^4}$ for $k> 5$ We have $\color{blue}{5k^4 \le \color{green}k\times k^4} \text{As } k> 5 $
Combining the above results we have
$\color{blue}{1< 5k < 10k^2 < 10k^3< 5k^4 < k^5} \text{ is true for} k > 5 \implies \text{ also true for } k > 6$
$$\begin{align*}
(k+1)^5 
& = k^5 + 5k^4 + 10k^3 + 10k^2 + 5k + 1\\
& < k^5 + k^5 + k^5 + k^5 + k^5 = 5k^5 < 5^{k+1}
\text{thus } P(k + 1) \text{is True}
\end{align*}$$
Now, if True of $P(k)$ $\implies $ True of $P(k+ 1) $ for $k> 6$ then $P(N>6) $ is True.
