$n^8 \lt 8^n-1$

Question

How can I prove that $n^8 \lt 8^n-1$ for all $n$?

I'm searching for technical way to show that without using calculus theorems, using well known facts such as $n^2 \lt 2^n$ and so.

Thanks!

Answer 1

While this fails for some $n$, such as $n=8$, exponential functions with base larger than $1$ always grow faster than polynomials. Therefore you should expect the inequality to hold for sufficiently large $n$. In fact, you can check that it holds for $n=9$, and if you're concerned only with integer solutions you can show by induction that it holds for all larger $n$.

If $n^8<8^n-1$ for some $n\geq 9$, then

$\begin{align*} 8^{n+1}-1&\gt 8(8^n-1)\gt 8n^8\\ &=(n+1)^8\left(8\left(\frac{n}{n+1}\right)^8\right)\\ &\geq (n+1)^8\left(8\left(\frac{9}{10}\right)^8\right)\\ &\gt (n+1)^8 . \end{align*}$

A way to think about why the right-hand side eventually dominates is to note that the ratio of successive terms on the left-hand side approaches $1$ (because $\frac{n+1}{n}=1+\frac{1}{n}$ approaches $1$), while the ratio of successive terms on the right-hand side is approximately $8$. This indicates that eventually the right-hand side grows much faster than the left-hand side.

Answer 2

It is simple to prove powers grow faster than polynomials, i.e. $\rm\ c > 1\ \Rightarrow\ c^n > p(n)\ $ for $\rm\ n > n_0\:.\:$ $\:$ For such it suffices to show eventually $\rm\ f(n) = c^n/p(n) > 1\:,\: $ or eventually $\rm\ f(n+1)/f(n) > 1 \ $ since, by multiplicative telescopy, $\rm\:f(n)\:$ is a product of these adjacent term ratios, viz.

$$\rm f(0)\ \ \prod_{k\:=\:0}^{n-1}\ \frac{f(k+1)}{f(k)}\ \ = \ \ \ \color{red}{\rlap{--}f(0)}\frac{\color{green}{\rlap{--}f(1)}}{\color{red}{\rlap{--}f(0)}}\frac{\color{royalblue}{\rlap{--}f(2)}}{\color{green}{\rlap{--}f(1)}}\frac{\phantom{\rlap{--}f(3)}}{\color{royalblue}{\rlap{--}f(2)}}\ \ \cdots\ \ \frac{\color{brown}{\rlap{--}f(n-1)}}{\phantom{\rlap{--}f(n-2)}}\frac{f(n)}{\color{brown}{\rlap{----}f(n-1)}}\ =\ \ f(n) $$

If you view the above cancellation as dynamically collapsing together the cancelling adjacent terms, then it is analogous to collapsing the tubes of handheld telescope into a single tube.

But $\rm\ f(k+1)/f(k)\ =\ c\ p(k)/p(k+1)\ \to\ c > 1\ $ as $\rm\ k\to \infty\ $ since for equal degree polynomials $\rm p(x),q(x)\:$ we have $\rm\ lim_{x\to\infty}\: p(x)/q(x) = p_0/q_0 = $ quotient of their leading coefficients, viz.

$$\rm \frac{p_0\ x^d + p_1\ x^{d-1}+\:\cdots\:+p_d}{q_0\ x^d + q_1\ x^{d-1}+\:\cdots\:+q_d}\ =\ \frac{p_0\ + p_1/x +\:\cdots\:+p_d/x^d}{q_0\ + q_1/x +\:\cdots\:+q_d/x^d}\ \to\ \frac{p_0}{q_0}\ \ \ as\ \ \ x\to\infty$$

As I have mentioned here before in many posts, by means of cancelling complicated expressions, telescopy often reduces induction problems to trivialities (here the fact that a product of terms eventually $> 1$ is itself eventually $> 1$). For example, the inductive proof in the answer by Jonas is essentially a telescopic proof of this type. Difficult problems involving hyperrational functions (i.e. $\rm\ f(n+1)/f(n) = $ rational function of $\rm\:n\:,\:$ such as powers and exponentials) are, after application of telescopy, greatly simplified to trivial problems about rational functions - functions so simple that questions about such can be decided mechanically by algorithms.

Answer 3

Let n=1, then you get 1<8-1. QED.