Partition of natural number not equal to factorial I wish to prove the following statement so I can use it as a lemma for a group theory result. To be honest I have not tried much yet, my intuition tells me this is going to be connected to the partition function or Euler's totient function, both of which are beyond my knowledge of number theory.

Prove that the following cannot happen,
$$\sum_{i=1}^{k} a_{i} = d \text{ and } \text{LCM}(a_{1},a_{2},...,a_{k}) = d!$$
$k, a_{i}, d \in \mathbb{N}$. $1 \leq a_{i} \leq d$. $a_{i}$'s not
  distinct. d > 2.

Any advice or comments greatly appreciated!

update:

As mentioned in the comments, intent is to use this to show a symmetric group is not cyclic. Also, here is a possible re-structuring and an outline of my thoughts..

It will suffice to show for $a_{i}, d \in \mathbb{N}$ s.t $1 \leq
 a_{i} \leq d$ and $2 < d$
$$\text{If } \text{LCM}(a_{1},a_{2},...,a_{k}) = d!, \text{ then }
 \sum_{i=1}^{k} a_{i} \neq d \text{ or } k > d$$

Idea: since $a_{i}'s < d$ except for when $a_{i} = d$ and $k = 1$, then it would imply that every number less than d was co prime which I can bring to a contradiction.
 A: Assume that $d\gt2$ and $\text{LCM}(a_1,a_2,\dots,a_k)=d!.$ We want to show that $a_1+a_2+\cdots+a_k\gt d.$
Let $p_1,p_2,\dots,p_n$ be the distinct prime divisors of $d!.$ From the fact that each $p_i$ divides some $a_j$ it easily follows that $a_1+\cdots+a_k\ge p_1+\cdots+p_n.$ So it will suffice to show that, for $d\gt2,$ the sum of the prime numbers $\le d$ is greater than $d.$ This can be proved by induction on $d,$ using Bertrand's postulate, which says that for $m\gt1$ there is a prime between $m$ and $2m.$
A: You want to show that
$\sum_{i=1}^{k} a_{i} = d 
\text{ and } \text{LCM}(a_{1},a_{2},...,a_{k}) = d!
$
can not happen.
For $k=1$,
$a_1 = d$
so this can and does happen
for $d \le 2$.
It does not happen for
$d \ge 3$.
For $k=2$,
you want
$lcm(a, d-a) = d!$
to not be.
The lcm is largest
when $a$ and $d-a$
are relatively prime,
so that
$lcm(a, d-a)
= a(d-a)
\le d^2/4
$.
Since
$d^2/4 < d!
\iff d/4 < (d-1)!
$
which is true for
$d \ge 1$,
your hope holds.
My next idea
would try a proof by induction
on $k$,
but it is late
and I am tired,
so I will leave it at this.
