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I found something that I was doing about 6 months ago that lacked substance in how I derived it, which is not my greatest trait when it comes to proof seeking.

Defining the following identity:

$$n^m+1-\Biggl\lfloor\Bigl(n^m+1\Bigr)^{\frac{1}{k}}\Biggr\rfloor^k=\prod^{\omega_{n,m,k}}_{j=1}p_{n,m,k,j}^{v_{n,m,k,j}}$$

$$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\text{(0)}$$

Implies the following inequality relation*: $$1 \lt n \lt N \,\land\,n \geq m \geq k \Rightarrow \omega_{n,m,k} \leq \frac{N}{2} $$ $$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\text{(1)}$$

For the case of $n=m=k \land n \gt1$ the inequality relation $\text{(1)}$ is held by a proof for the following identity, for which I am seeking hints as to what theorem / congruence relation I would use in such a proof: $$n^n+1-\Biggl\lfloor\Bigl(n^n+1\Bigr)^{\frac{1}{n}}\Biggr\rfloor^n=1 \quad\forall n \gt 1$$ $$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\text{(2)}$$

But this is only a case, which by no means proves the inequality relation $\text{(1)}$ overall. And because I didn't provide any other derivation of it, it's safe to assume that it was arrived at intuitively, which, as helpful as it can be, is synonymous with flawed logic. So in that regard I need to post it now, to provide others the opportunity to find a counter example, as it would eliminate me spending time seeking proof for a false statement.

Thanks in advance.

Edit: I will link the following posts of mine with obvious relevance in order of priority as I currently see it.

One

*Uncertainty regarding whether or not the implicative arrow is bijective or not

The following plots are provided for the reader to note the highest density of points in a given range occurs when $\omega_{n,m,k} =3$:

enter image description here enter image description here enter image description here enter image description here

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If I understand your question correctly, you're asking for several things, including a proof of your (2), i.e.,

$$n^n+1-\Biggl\lfloor\Bigl(n^n+1\Bigr)^{\frac{1}{n}}\Biggr\rfloor^n=1 \quad\forall \; n \gt 1 \tag{2}\label{eq2A}$$

If so (and, if not, please let me know so I can delete this answer), then to prove \eqref{eq2A}, note that

$$\Bigl(n^n+1\Bigr)^{\frac{1}{n}} \gt \Bigl(n^n\Bigr)^{\frac{1}{n}} = n \tag{3}\label{eq3A}$$

Also, by the Binomial Theorem, you have

$$m = \left(n+1\right)^n = n^n + \binom{n}{1}n^{n-1} + \ldots + \binom{n}{n-1}n + 1 \gt n^n + 1 \tag{4}\label{eq4A}$$

Thus, you also have

$$m^{\frac{1}{n}} = n + 1 \gt \Bigl(n^n+1\Bigr)^{\frac{1}{n}} \tag{5}\label{eq5A}$$

Using \eqref{eq3A} and \eqref{eq5A} together shows

$$n + 1 \gt \Bigl(n^n+1\Bigr)^{\frac{1}{n}} \gt n \implies \Biggl\lfloor\Bigl(n^n+1\Bigr)^{\frac{1}{n}}\Biggr\rfloor = n \tag{6}\label{eq6A}$$

Substituting this into the LHS of \eqref{eq2A} gives

$$n^n+1-\Biggl\lfloor\Bigl(n^n+1\Bigr)^{\frac{1}{n}}\Biggr\rfloor^n = n^n + 1 - n^n = 1 \tag{7}\label{eq7A}$$

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  • $\begingroup$ No please don't delete this as you were correct in interpreting my question as two questions, you very well could have concisely answered the second of which, I just felt it not in good form to post them seperately $\endgroup$ – Adam Oct 22 '19 at 0:22
  • $\begingroup$ is (5) meant to be an n for the base rather than m on extreme lhs as a typo? $\endgroup$ – Adam Oct 22 '19 at 0:24
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    $\begingroup$ @Adam Thanks for confirming what you were asking for. As for my (5), no it's not a typo. The $m$ comes from (4), where $m = \left(n + 1\right)^n$, which is how I got $m^{\frac{1}{n}} = n + 1$. I was doing this to show you can bound the value within the floor function between $n$ and $n+1$, showing the result of the floor function will always be $n$. $\endgroup$ – John Omielan Oct 22 '19 at 0:26
  • $\begingroup$ sorry (4) and (5) I meant I am not following with the introduction of $m$, but it is first glance and you answer is very much in the directions I have attempted thus far, so please appreciate that I will take some time to accept, in addition to the absence of the proof of the inequality in my original $\endgroup$ – Adam Oct 22 '19 at 0:27
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    $\begingroup$ @Adam I introduced $m$ to avoid having to use $2$ levels of exponents, i.e., $\frac{1}{n}$ and $n$, anywhere in the solution. I'm sorry if that made what I was doing more confusing. As for a proof of the inequality in your original, that's likely to be quite a bit more complicated. Regarding not posting them separately, I would suggest it better to do that, but make sure you link the $2$ to see how they're connected. $\endgroup$ – John Omielan Oct 22 '19 at 0:28

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