How can I show $1-\frac{1}{x}+x^{1-\frac{1}{x}}1$? Denote $$f(x):=1-\frac{1}{x}+x^{1-\frac{1}{x}}$$

How can I prove that $f(x)<x$ holds for every real $x>1$ ?

Wolfram gives the taylor series 
$$f(x)=1+(x-1)-\frac{1}{2}(x-1)^3+\frac{4}{3}(x-1)^4-\frac{31}{12}(x-1)^5+O((x-1)^6)$$
But I would like to have a proof without the taylor series because it is difficult to estimate the remainder theorem by hand. The Lambert-W function or logarithming $x^{1-\frac{1}{x}}$ might help, but I did not succeed with either of these methods.
 A: Letting $u=\frac1x$,
$$
1-\frac1x+x^{1-\frac1x}\lt x\tag{1}
$$
for $x\gt1$ is, the same as
$$
u^u\lt1-u(1-u)\tag{2}
$$
for $0\lt u\lt1$.

Since $\frac{\log(1-v)}v$ is monotonically decreasing for $0\lt v\lt1$, we get
$$
\frac{\log(1-v)}v\lt\frac{\log(1-v(1-v))}{v(1-v)}\tag{3}
$$
multiplying both sides of $(3)$ by $v(1-v)$ gives
$$
(1-v)\log(1-v)\lt\log(1-v(1-v))\tag{4}
$$
Substituting $u=1-v$ in $(4)$ gives the logarithm of $(2)$.

Details of the Monotonicity of $\boldsymbol{\frac{\log(1-v)}v}$
Since $e^x\ge1+x$, with equality only when $x=0$, we get $e^{-x}\ge1-x$ and $e^x\le\frac1{1-x}$, still with equality only when $x=0$. Substitute $x\mapsto-\frac v{1-v}$ and take logarithms:
$$
-\frac v{1-v}\le\log(1-v)\tag{5}
$$
with equality only when $v=0$. Therefore, for $0\lt v\lt1$,
$$
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}v}\frac{\log(1-v)}v
&=\frac{-\frac v{1-v}-\log(1-v)}{v^2}\\
&\lt0\tag{6}
\end{align}
$$
A: It is sufficient to prove that
$$\log\left(x-1+\frac 1x\right)-\left(1-\frac 1x\right)\log(x)\gt 0$$
for $x\gt 1$.
Let $$g(x):=\log\left(x-1+\frac 1x\right)-\left(1-\frac 1x\right)\log(x)$$
Then,
$$g'(x)=\frac{2x^2-3x+1+(-x^2+x-1)\log(x)}{x^2 (x^2-x+1)}$$
Let $h(x):=2x^2-3x+1+(-x^2+x-1)\log(x)$. Then,
$$h'(x)= 3 x-\frac 1x-2 x \log(x)+\log(x)-2$$
$$h''(x)=\frac{1}{x^2}+\frac{1}{x}-2 \log(x)+1$$
$$h'''(x)=-\frac{2 x^2+x+2}{x^3}\lt 0$$
Hence, $h''(x)$ is decreasing with $h''(1)=3$ and $\lim_{x\to \infty}h''(x)=-\infty$. So, there exists only one $x=\alpha\gt 1$ such that $h''(\alpha)=0$. So, $h'(x)$ is increasing for $1\lt x\lt\alpha$ and is decreasing for $x\gt \alpha$ with $h'(1)=0$ and $\lim_{x\to\infty}h'(x)=-\infty$. So, there exists only one $x=\beta\gt 1$ such that $h'(\beta)=0$. So, $h(x)$ is increasing for $1\lt x\lt\beta$ and is decreasing for $x\gt\beta$ with $h(1)=0$ and $\lim_{x\to\infty}h(x)=-\infty$. Therefore, there exists only one $x=\gamma\gt 1$ such that $g'(\gamma)=0$. So, $g(x)$ is increasing for $1\lt x\lt\gamma$ and is decreasing for $x\gt \gamma$ with $g(1)=0$ and $\lim_{x\to \infty}g(x)=0$. 
It follows from these that $g(x)\gt 0$.
