$x>1$and $k>0$ implies $x^k>1$ If $x>1$ and $k>0$,does it imply that $x^k>1$ ?It seems like a very trivial result;however i cannot seem to show it analytically.
Analytic proof would be appreciated.
 A: The power $x^k$, for $x>0$, can be defined/computed as
$$
x^k=\exp(k\log x)
$$
where $\exp$ and $\log$ are the exponential and logarithmic function on base $e$.
Since $x>1$, we have $\log x>0$, so $k\log x>0$ and $\exp(k\log x)>1$.
The same relation as above shows, with the chain rule, that the derivative of $f(x)=x^k=\exp(k\log x)$ is
$$
f'(x)=\exp(k\log x)\frac{k}{x}=k\frac{x^k}{x}
$$
which is positive, so the function is increasing. Since $f(1)=1$, we reach the conclusion with a different method.
A: Just try:
Assume $x^{k}$$\le$1,so $\frac{1}{x^{k}}$$\ge$1,so $(1/x)^{k}$$\ge$$1^{k}$
So,(1/x)$\ge1$ implies $x\le1$ which is contradiction.
A: It depends on what framework you want to prove it.
Let $f(x) = x^k$ then $f'(x) = kx^{k-1}$ and $x^{k-1} > 0$ so $f(x)$ is increasing if $x > 0$. So as $f(1) = 1$, $f(x > 1) > 1$.
Or let $f(k) = x^k$ (where $x$ is constant) then $f'(x) = e^{k\ln x} = (\ln x )x^k$.  As $x > 1$, $\ln x > 0$ and $x^k > 0$ so $f(x)$ is increasing.  So $f(0) = 1$ so $f(k > 0)  >1$.
But that's probably defeating the purpose, as I imagine you are probably trying to go back to basics and haven't analytically determined these rules of calculus yet?
Which means this depends on how $x^k$ was defined.
If $k \in \mathbb Z$ then this follows inductively as $x^{k-1}= x^k*\frac 1x < x^k < x^k *x = x^{k-1}$.  So if $k>0$ or $k \ge 1$ then $x^k > x^0 = 1$.
If $k \in \mathbb Q$ then $x^{k=\frac nm; n\ge 1;m\ge 1} \le 1\implies (x^{\frac nm})^m \le 1 \implies x^n \le 1$ but as we showed above that is impossible.
If $k \in \mathbb R$ and you have defined $x^k$ where $k = \lim q_n; q_n \in \mathbb Q$ to be $\lim x^{q_n}$ then it follows as a standard limit proof.  ($\lim q_n > 0$ so there is an $N$ so that $n > N\implies q_n > 0$ so $x^{q_n} > 1$ so $\lim x^{q_n} > 1$).
The standard definition is that $x^k = e^{k\ln x}$ where $e = \lim(1 + \frac 1n)^n$ and $\ln x = \int_{1}^\infty \frac 1x dx$ in which case we can probably use the arguments at the beginning of this post.
A: $f(x)=x^k$, $k,x$ real, positive.
$f(1)=1$. $f'(x)= kx^{k-1} >0$, implies
$f(x)$ is increasing, I.e.
$f(x) > 1$ for $x> 1.$
