Note that the $10^{k}$'s place of $a_{n}$ is 1's place of $\left\lfloor n^{\frac{2018}{2017}}/10^{k}\right\rfloor $.If $\left(n+1\right)^{\frac{2018}{2017}}/10^{k}-n^{\frac{2018}{2017}}/10^{k}\leq1$,then $10^{k}$'s place of $a_{n},a_{n+1}$ don't skip, i.e. they are same or consective. Therefore, if 1. $\left(n+N\right)^{\frac{2018}{2017}}/10^{k}-n^{\frac{2018}{2017}}/10^{k}\geq10$ 2. $\left(m+1\right)^{\frac{2018}{2017}}/10^{k}-m^{\frac{2018}{2017}}/10^{k}\leq1$ for any $m$ with $n\leq m\leq n+N-1$ we have $a_{n},a_{n+1},\ldots,a_{n+N-1}$ contains all digits from 0 to 9 in $10^{k}$'s place. We find such $N$ satisfing these constraints. Let $f\left(x\right)=x^{\frac{2018}{2017}}$. Since $\left(m+1\right)^{\frac{2018}{2017}}-m^{\frac{2018}{2017}}\leq f\left(m+1\right)-f\left(m\right)\leq f'\left(m+1\right)=\frac{2018}{2017}\left(m+1\right)^{\frac{1}{2017}}\leq\frac{2018}{2017}\left(n+N\right)^{\frac{1}{2017}}$ for $n\leq m\leq n+N-1$. Thus, $\frac{2018}{2017}\left(n+N\right)^{\frac{1}{2017}}\leq10^{k}$ is sufficient to satisfy 2. . It is equivalent to $\log_{10}\left(\frac{2018}{2017}\right)+\frac{1}{2017}\log_{10}\left(n+N\right)\leq k$. Therefore, we can take integer $k$ such that $k\leq\log_{10}\left(\frac{2018}{2017}\right)+\frac{1}{2017}\log_{10}\left(n+N\right)+1$.Then, we have \begin{align*} 10^{k+1} & =10\cdot10^{k}\\ & \leq10\cdot10\cdot\frac{2018}{2017}\left(n+N\right)^{\frac{1}{2017}}. \end{align*}On the other hand, from Bernouli's inequality, we have \begin{align*} \left(n+N\right)^{\frac{2018}{2017}}-n^{\frac{2018}{2017}} & \geq n^{\frac{2018}{2017}}\left(\left(1+\frac{N}{n}\right)^{\frac{2018}{2017}}-1\right)\\ & \geq n^{\frac{2018}{2017}}\left(\frac{2018}{2017}\cdot\frac{N}{n}+1-1\right)\\ & =n^{\frac{1}{2017}}\cdot\frac{2018}{2017}\cdot N \end{align*}Therefore, to satisfy 1. , \begin{align*} n^{\frac{1}{2017}}\cdot\frac{2018}{2017}\cdot N & \geq10\cdot10\cdot\frac{2018}{2017}\left(n+N\right)^{\frac{1}{2017}}\\ n^{\frac{1}{2017}}N & \geq10^{2}\left(n+N\right)^{\frac{1}{2017}}\\ nN^{2017} & \geq10^{2\cdot2017}\left(n+N\right) \end{align*}is sufficient. We must find $N$ such that this is true for any positive integer $n$. Actually, we can take $N=10^{3}$. Then, we have \begin{align*} nN^{2017}-10^{2\cdot2017}\left(n+N\right) & =10^{3\cdot2017}n-10^{2\cdot2017}n-10^{2\cdot2017+3}\\ & \geq\left(10^{2\cdot2017}+10^{2\cdot2017+3}\right)n-10^{2\cdot2017}n-10^{2\cdot2017+3}\\ & =10^{2\cdot2017+3}\left(n-1\right)\\ & \geq0. \end{align*}

We will prove that some term among $a_m,\ldots,a_{m+999}$ will contain the digit $5$ for all sufficiently large $m$, which clearly finishes. By Newton's extended binomial formula, we have $$d_k:=(k+1)^{\tfrac{2018}{2017}}-k^{\tfrac{2018}{2017}}=\frac{2018}{2017}k^{\tfrac{1}{2017}}+o(1).$$Hence pick $m$ large enough such that the $o(1)$ terms are all less than $\tfrac{1}{1000}$. Furthermore, since $(x+999)^{\tfrac{1}{2017}}-x^{\tfrac{1}{2017}}$ is decreasing for $x \in \mathbb{R}^+$ and has limit $0$, pick $m$ large enough such that $(m+999)^{\tfrac{1}{2017}}-m^{\tfrac{1}{2017}}<\tfrac{1}{1000}\cdot \tfrac{2017}{2018}$. Then the difference between any two $d_{m+i}$ for $0 \leq i \leq 999$ is at most $\tfrac{1}{500}$. For any reals $x,y$, we have $\lfloor x+y\rfloor - \lfloor x \rfloor \in \{\lfloor y \rfloor, \lceil y \rceil\}$. Therefore the difference $a_{m+i+1}-a_{m+i}$ (where $0 \leq i \leq 999$) is either $\lfloor d_{m+i} \rfloor$ or $\lceil d_{m+i} \rceil$. Because each $d_{m+i}$ is within $\tfrac{1}{500}$ of all the others, the set $\{c-2,c-1,c\}$ for some integer $c$ will contain all the possible values of $a_{m+i+1}-a_{m+i}$. By taking $m$ to be sufficiently large, suppose that $c \geq 100$. Suppose that $c$ has $k\geq 3$ digits; I will prove that some $a_{m+i}$ has $5$ as its $k+1$-th digit. We have $$a_{m+999}-a_m=(a_{m+999}-a_{m+998})+\cdots+(a_{m+1}-a_m) \geq 998(c-2)\geq 998(10^{k-1}-2)\geq 10^{k+1},$$but since $\{c-2,c-1,c\}$ are all less than $10^k$, for discrete continuity reasons every $k+1$-th digit must be observed amongst $a_m,\ldots,a_{m+999}$, since we must "pass through" that digit because $a_{m+999}-a_m\geq 10^{k+1}$, but we cannot "skip" it because each individual difference is too small. Thus we will certainly have some term of this sequence containing $5$ as a digit, as desired. $\blacksquare$ Remark: The bound $N=1000$ is loose; something like $N=102$ should work as well for sufficiently large $m$.