Find $a>0$ if the floor of $(n^{2}-n)(\sqrt[n]{a}-1)$ is equal to $n-1$. Find $a>0$ knowing that for every n non-zero natural number, the floor of $(n^{2}-n)(\sqrt[n]{a}-1)$ is equal to $n-1$.
I know a is e, because taking the limit of the expresion we find that a is e. But i dont know how to prove that this works for every n. I am thniking at induction, but i dont know how i can use it. Can someone help me solve this problem in a logic form.

We have that $$ \lfloor (n^2 - n)(a^{1/n} - 1)\rfloor = n-1$$
Since $x-1< \lfloor x\rfloor \le x,$ we can conclude that for every $n$, $$ n -1 \le (n^2-n)(a^{1/n} - 1) \le n \\ \iff \left(1 + \frac{1}{n}\right)^n \le a \le \left(1 + \frac{1}{n-1}\right)^n.$$ Now taking the limit shows that if $a$ exists then $a = e$.
So, the only possibility for $a$ is $e$, but we need to verify that this works. For $n = 1,$ this is obvious. For $n \ge 2,$ the following fact can be exploited: $$1 + \frac{1}{n} < e^{1/n} < 1 + \frac{1}{n-1}. $$ Note the strict inequalities. This does the job, since the relations imply
$$ (n^2 - n)(e^{1/n} - 1)< n \implies \lfloor (n^2 - n) (e^{1/n} - 1)\rfloor \le n-1, \\ (n^2 - n)(e^{1/n} - 1) > n-1 \implies \lfloor (n^2 - n) (e^{1/n} - 1)\rfloor \ge n-1,$$ forcing the floor in question to be $n-1$.
It remains to show the bounds I posited. The lower bound is trivial (truncate the Taylor series). For the upper bound, it suffices to show that for $n \ge 2,$ $$ \sum_{k \ge 2} \frac{1}{k!n^k} < \frac{1}{n(n-1)}$$
But notice that for $k \ge 4,$ $$ \frac{1}{k! n^k} \le \frac{1}{(2n)^k}.$$ Summing the geometric series, and using $n \ge 2,$ we have that $$ \sum \frac{1}{k! n^k} \le \frac{1}{2n^2} + \frac{1}{6n^3} + \frac{1}{2^4n^4} \cdot \frac{1}{1 - 1/2n}\\\le  \frac{1}{2n^2} + \frac{1}{6n^2} + \frac{1}{2^4n^4} \cdot \frac{4}{3} \\ \le  \frac{1}{2n^2} + \frac{1}{12n^2} + \frac{1}{48n^2 } <\frac{29}{48n^2} < \frac{5}{8n(n-1)},$$
and we're done.
