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Wow! I have been thinking about this problem, too. I can prove that for every odd natural number $d$ there exists infinitely many $n \in \mathbb{N}$ such that $n^{d}+1 \mid n!$. Here you have my gorgeous proof of the case $d=3$ of the result (cf. https://oeis.org/A270441):

For every $k \in \mathbb{N}$ we have that

\begin{eqnarray*}(k^{6}+2k^{4}+k^{2}+1)^{3}+1 &=& (k^{2}-k+1)(k^{2}+2)(k^{2}+k+1) &\cdot& (k^{4}-k^{3}+2k^{2}-2k+1)(k^{4}+1)(k^{4}+k^{3}+2k^{2}+2k+1);\end{eqnarray*}

therefore, it follows that, for every $k \in \mathbb{N} \setminus \{1\}$, the natural number $N_{k}:=k^{6}+2k^{4}+k^{2}+1$ is such that $N_{k}^{3}+1$ divides $N_{k}!$. QED.

I have to finish something else now, but if you are interested in what I have mentioned, I may add more details about the result which I obtained later on.

Wow! I have been thinking about this problem, too. I can prove that for every odd natural number $d$ there exists infinitely many $n \in \mathbb{N}$ such that $n^{d}+1 \mid n!$. Here you have my gorgeous proof of the case $d=3$ of the result (cf. https://oeis.org/A270441):

For every $k \in \mathbb{N}$ we have that

\begin{eqnarray*}(k^{6}+2k^{4}+k^{2}+1)^{3}+1 &=& (k^{2}-k+1)(k^{2}+2)(k^{2}+k+1) &\cdot& (k^{4}-k^{3}+2k^{2}-2k+1)(k^{4}+1)(k^{4}+k^{3}+2k^{2}+2k+1);\end{eqnarray*}

therefore, it follows that, for every $k \in \mathbb{N} \setminus \{1\}$, the natural number $N_{k}:=k^{6}+2k^{4}+k^{2}+1$ is such that $N_{k}^{3}+1$ divides $N_{k}!$. QED.

I have to finish something else now, but if you are interested in what I have mentioned, I may add more details about the result that I obtained later on.

Wow! I have been thinking about this problem, too. I can prove that for every odd natural number $d$ there exists infinitely many $n \in \mathbb{N}$ such that $n^{d}+1 \mid n!$. Here you have my gorgeous proof of the case $d=3$ of the result (cf. https://oeis.org/A270441):

For every $k \in \mathbb{N}$ we have that

\begin{eqnarray*}(k^{6}+2k^{4}+k^{2}+1)^{3}+1 &=& (k^{2}-k+1)(k^{2}+2)(k^{2}+k+1) &\cdot& (k^{4}-k^{3}+2k^{2}-2k+1)(k^{4}+1)(k^{4}+k^{3}+2k^{2}+2k+1);\end{eqnarray*}

therefore, it follows that, for every $k \in \mathbb{N} \setminus \{1\}$, the natural number $N_{k}:=k^{6}+2k^{4}+k^{2}+1$ divides $N_{k}!$. QED.

I have to finish something else now, but if you are interested in what I have mentioned, I may add more details about the result which I obtained later on.

Wow! I have been thinking about this problem, too. I can prove that for every odd natural number $d$ there exists infinitely many $n \in \mathbb{N}$ such that $n^{d}+1 \mid n!$. Here you have my gorgeous proof of the case $d=3$ of the result (cf. https://oeis.org/A270441):

For every $k \in \mathbb{N}$ we have that

\begin{eqnarray*}(k^{6}+2k^{4}+k^{2}+1)^{3}+1 &=& (k^{2}-k+1)(k^{2}+2)(k^{2}+k+1) &\cdot& (k^{4}-k^{3}+2k^{2}-2k+1)(k^{4}+1)(k^{4}+k^{3}+2k^{2}+2k+1);\end{eqnarray*}

therefore, it follows that, for every $k \in \mathbb{N} \setminus \{1\}$, the natural number $N_{k}:=k^{6}+2k^{4}+k^{2}+1$ is such that $N_{k}^{3}+1$ divides $N_{k}!$. QED.

I have to finish something else now, but if you are interested in what I have mentioned, I may add more details about the result which I obtained later on.

Wow! I have been thinking about this problem, too. I can prove that for every odd natural number $k$ there exists infinitely many $n \in \mathbb{N}$ such that $n^{k}+1 \mid n!$. Here you have my gorgeous proof of the case $k=3$ of the result (cf. https://oeis.org/A270441):

For every $k \in \mathbb{N}$ we have that

\begin{eqnarray*}(k^{6}+2k^{4}+k^{2}+1)^{3}+1 &=& (k^{2}-k+1)(k^{2}+2)(k^{2}+k+1) &\cdot& (k^{4}-k^{3}+2k^{2}-2k+1)(k^{4}+1)(k^{4}+k^{3}+2k^{2}+2k+1);\end{eqnarray*}

therefore, it follows that, for every $k \in \mathbb{N} \setminus \{1\}$, the natural number $N_{k}:=k^{6}+2k^{4}+k^{2}+1$ divides $N_{k}!$. QED.

I have to finish something else now, but if you are interested in what I have mentioned, I may add more details about the result which I obtained later on.

Wow! I have been thinking about this problem, too. I can prove that for every odd natural number $d$ there exists infinitely many $n \in \mathbb{N}$ such that $n^{d}+1 \mid n!$. Here you have my gorgeous proof of the case $d=3$ of the result (cf. https://oeis.org/A270441):

For every $k \in \mathbb{N}$ we have that

\begin{eqnarray*}(k^{6}+2k^{4}+k^{2}+1)^{3}+1 &=& (k^{2}-k+1)(k^{2}+2)(k^{2}+k+1) &\cdot& (k^{4}-k^{3}+2k^{2}-2k+1)(k^{4}+1)(k^{4}+k^{3}+2k^{2}+2k+1);\end{eqnarray*}

therefore, it follows that, for every $k \in \mathbb{N} \setminus \{1\}$, the natural number $N_{k}:=k^{6}+2k^{4}+k^{2}+1$ divides $N_{k}!$. QED.

I have to finish something else now, but if you are interested in what I have mentioned, I may add more details about the result which I obtained later on.