Fix $k$ arbitrarily and denote the sum $k! + (2k)! + \cdots + (nk)!$ by $I_n$. It is enough to show that the set of primes $p$ that divide at least one of the $I_n$ is infinite—here is an argument that this is so.
Suppose otherwise that there are only finitely many primes that divide any of the $I_n$. For a given prime $p$ and integer $a$, let $e_p(a)$ be the exponent of $p$ in $a$. We'll need the following property of $e_p$: for any pair $a,b$ of integers,
\begin{align*}
e_p(a+b) = \min\{e_p(a),e_p(b)\} \tag{1}
\end{align*}
except possibly when $e_p(a) = e_p(b)$.
Let $U$ be the set of all primes $p$ such that $e_p(I_{n-1}) \geq e_p((nk)!)$ for all $n$ (the primes with unbounded exponent in the sequence $\{I_n\}$), and let $B$ be the complementary set, i.e., the set of all primes $p$ such that $e_p(I_{n-1})<e_p((nk)!)$ for some $n$. Note that if $e_p(I_{n-1}) < e_p((nk)!)$ then $e_p(I_{n-1}) = e_p(I_n) = \cdots$ by virtue of $(1)$. Since there are only finitely many primes $p \in B$ for which $e_p(I_n)$ is ever nonzero, there is an integer $N$ with the property that $e_p(I_n) = e_p(I_N)$ for each $p \in B$ and for all $n \geq N$ (any $N$ such that $Nk$ is at least as big as the largest prime that divides some $I_n$ will do). The point is that $e_p(I_n)$ is constant for sufficiently large $n$ unless $p\in U$.
We have required only that $e_p(I_{n-1}) \geq e_p((nk)!)$ when $p \in U$. But $(1)$ implies that in certain circumstances the two exponents must be equal. Indeed, where $((n+1)k)!$ contains more factors of $p$ than $(nk)!$, we must have $e_p(I_{n-1}) = e_p((nk)!)$ lest $(1)$ give
\begin{align*}
e_p(I_n) = \min\{e_p(I_{n-1}),e_p((nk)!)\} = e_p((nk)!) < e_p(((n+1)k)!)
\end{align*}
in violation of the assumption that $p \in U$.
If we now choose $n$ carefully we can force $e_p(I_{n-1}) = e_p((nk)!)$ for all $p \in U$. The foregoing argument shows that this will happen whenever $e_p((nk)!) < e_p(((n+1)k)!)$ for all $p\in U$. If $m = \prod_{p\in U}p$, then $n = am -1$ satisfies this condition for all $a\in\mathbb{N}$. Thus:
\begin{align*}
e_p(I_{am-2}) = e_p(((am-1)k)!) \quad\text{for $p\in U$ and $a\in\mathbb{N}$.}
\end{align*}
We just need one more fact, and that is that $((am-1)k)!$ contains no more factors of any prime $p \in U$ than $((am-2)k)!$ does. If this is true, then we get the following chain for $p\in U$ and $a\in\mathbb{N}$:
\begin{align*}
e_p(I_{am-3})\geq e_p(((am-2)k)!) = e_p(((am-1)k)!) = e_p(I_{am-2}). \tag{2}
\end{align*}
Since for large $a$ we have $e_p(I_{am-3}) = e_p(I_{am-2})$ for $p \notin U$, the chain $(2)$ implies for large $a$ that $I_{am-3}$ contains at least as many factors of every prime $p$ as $I_{am-2}$ does. This in turn means that $I_{am-3}\geq I_{am-2}$, which is our desired contradiction.
So it remains only to show that
\begin{align*}
e_p(((am-1)k)!) = e_p(((am-2)k)!)\quad\text{for $p\in U$}. \tag{3}
\end{align*}
Since $((am-1)k)!/((am-2)k)! = (amk - k)(amk-k-1)\cdots (amk - 2k+1)$, the exponents will be unequal only if $p$ divides $amk-j$ for some $j<2k$. For $p\in U$ this will be true only if $p$ divides $j$, since in that case $p$ divides $m$. In particular, we would need $p<2k$. But no prime $p$ smaller than $2k$ has unbounded exponent in the sequence $\{I_n\}$ because $I_n/k!$ is not divisible by any such prime. Thus every prime $p\in U$ is bigger than $2k$, and $(3)$ is proved.
Consider $f(x,y) = x^y - y^x$ where $x, y \approx 3$. Increasing $x$ should bring $f$ down and increasing $y$ should bring $f$ up. In general, without any real knowledge of $f$:
$$ 1 > f(x+ \epsilon_1, y + \epsilon_2) \approx
f(x, y ) + \epsilon_1 \frac{\partial f}{\partial x} + \epsilon_2 \frac{\partial f}{\partial y}
$$
where the first inequality is something we are trying to prove for $x = e \approx 2.718$ and $y = \pi \approx 3.141$.
Let's try $x = \frac{11}{4}$ and $y = \frac{13}{4}$. Then we have taken off too much slack:
$$ \left( \frac{11}{4} \right)^{\frac{13}{4}} - \left( \frac{13}{4} \right)^{\frac{11}{4}} \approx 1.214$$
Using the continued fractions of $e$ and $\pi $ does give less than one and check with a calculator:
$$ \left( \frac{8}{3} \right)^{\frac{22}{7}} - \left( \frac{22}{7} \right)^{\frac{8}{3}} \approx 0.621$$
Even your proof requires a calculator in the last step... but we did not use the exact values of $\pi, e$.
How good is this approximation? Somehow we should compute how quickly $f$ changes with $x$ and $y$:
$$ \frac{\partial f}{\partial x} = y \, x^{y-1} - (\ln y ) y^x \approx 3 \times 3^{3-1} - (\ln 3) 3^3 \approx -27(\ln \frac{e}{3}) \approx -2.7 $$
The $y$ partial derivative is similar, except it is positive instead of negative.
It can be proven, the continued fraction error of $e$ and $\pi$ are inversely proportional to the denominators:
$$ \left| e - \frac{8}{3} \right| < \frac{1}{3^2} \text{ and } \left| \pi - \frac{22}{7} \right| < \frac{1}{7^2}$$
These would be our estimates of $\epsilon_1$ and $\epsilon_2$. Then using a tiny bit of multivariable calculus:
$$ \epsilon_1 \frac{\partial f}{\partial x} + \epsilon_2 \frac{\partial f}{\partial y}
\approx \frac{1}{3^2} \times 2.7 + \frac{1}{7^2} \times (-2.7) < 0.355 $$
This should be enough to establish $|f(\pi,e)| < 1$.
This used a calculator in many places but not the exact value of the constants $\pi$ and $e$.
Best Answer
$\Large {3^{3^4} \over 4^{4^3}} = {3^{81} \over 4^{64}} = {3 \over \left({256\over243}\right)^{16}} > {3 \over (1+{13\over243})^{243\over13}} > {3 \over e} > 1$