Evaluating $\lim_{x\to 1}\frac{^5x-^4x}{(1-x)^5}$, where $^nx$ is the repeated exponent ("tetration") operation 
$$\lim_{x\rightarrow1}\frac{x^{x^{x^{x^x}}}-{x^{x^{x^x}}}}{(1-x)^5}$$

My friends told that it appeared on Instagram (some social media network). I tried various methods but failed. (I tried using Taylor series but failed. Tried l'Hopital but failed. Tried substitution but failed. Gave up here.) They also said that some of the comments claim that the answer is $\;-1.$
 A: Let $f_0(x)=1$, $f_n(x) = x^{x^{x^{\cdots^{x}}}}=\, ^nx$ be $n$ iterated exponent. We can see
$$\begin{eqnarray}
\lim_{x\to 1}\frac{f_1(x)-f_0(x)}{x-1}=\lim_{x\to 0}\frac{x-1}{x-1}=1.
\end{eqnarray}$$ If we define $F(t,x)=t^x =e^{x\log t}$, we can see by mean value theorem that
$$\begin{eqnarray}
F(x,f_n(x))-F(x,f_{n-1}(x))&=&F_2\left(x,s_xf_n(x)+(1-s_x)f_{n-1}(x)\right)(f_n(x)-f_{n-1}(x))
\end{eqnarray}$$ for some $s_x\in (0,1)$. Assume $$\lim\limits_{x\to 1}\frac{f_n(x)-f_{n-1}(x)}{(x-1)^n}=a_n$$ exists. Then
$$\begin{eqnarray}
\lim\limits_{x\to 1}\frac{f_{n+1}(x)-f_{n}(x)}{(x-1)^{n+1}}&=&\lim\limits_{x\to 1}\frac{F(x,f_n(x))-F(x,f_{n-1}(x))}{(x-1)^{n+1}}\\
&=&\lim\limits_{x\to 1}\frac{F_2\left(x,s_xf_n(x)+(1-s_x)f_{n-1}(x)\right)(f_n(x)-f_{n-1}(x))}{(x-1)^{n+1}}\\&=&a_n\cdot \lim\limits_{x\to 1}\frac{F_2\left(x,s_xf_n(x)+(1-s_x)f_{n-1}(x)\right)}{x-1}\\&=&a_n\cdot \lim\limits_{x\to 1}\frac{\log x\cdot x^{s_xf_n(x)+(1-s_x)f_{n-1}(x)}}{x-1}\\
&=&a_n.
\end{eqnarray}$$ By induction, it follows that $a_{n+1}=a_n =\cdots =a_1 =1$. Hence
$$
\lim_{x\rightarrow1}\frac{x^{x^{x^{x^x}}}-{x^{x^{x^x}}}}{(1-x)^5}=-a_5=-1.
$$
A: Start using composition of Taylor series using mainly $t=e^{\log(t)}$. Then
$$x^x=1+(x-1)+(x-1)^2+\frac{1}{2} (x-1)^3+\frac{1}{3} (x-1)^4+\frac{1}{12}
   (x-1)^5+O\left((x-1)^6\right)$$
$$x^{x^x}=1+(x-1)+(x-1)^2+\frac{3}{2} (x-1)^3+\frac{4}{3} (x-1)^4+\frac{3}{2}
   (x-1)^5+O\left((x-1)^6\right)$$
$$x^{x^{x^x}}=1+(x-1)+(x-1)^2+\frac{3}{2} (x-1)^3+\frac{7}{3} (x-1)^4+3
   (x-1)^5+O\left((x-1)^6\right)$$
$$x^{x^{x^{x^x}}}=1+(x-1)+(x-1)^2+\frac{3}{2} (x-1)^3+\frac{7}{3} (x-1)^4+4
   (x-1)^5+O\left((x-1)^6\right)$$ Which shows the limit.
If you continue the expansion, you would get
$$\frac{x^{x^{x^{x^x}}}-x^{x^{x^x}}}{(1-x)^5}=-1-2 (x-1)-\frac{13}{3} (x-1)^2+O\left((x-1)^3\right)$$
Use your computer with $x=1.1$. The "exact" value would be $\approx -1.2525$ while the above expansion gives $-\frac{373}{300}\approx -1.2433$.
