Showing n! is greater than n to the tenth power I'd like to show $n!>n^{10} $ for large enough n ( namely $ n \geq 15 $). 
By induction, I do not know how to proceed at this step: 
$$ (n+1)\cdot n!>(n+1)^{10}  $$
As I can't see how to simplify $(n+1)^{10} $. 
This seems like such a trivial thing (and it probably is), yet I can't do it. Isn't there an easier way to show this? 
(P.S. I need to refrain from the use of derivatives, integrals etc., I suppose, then you could work something out with the slope of the respective functions)
 A: Remember to show the base case, namely that $15! > 15^{10}$. After that is verified, make the induction hypothesis (IH):
IH: $m! > m^{10}$
Show: $(m+1)! > (m+1)^{10}$ using IH.
As you've done so far, what we need to show is:
\begin{align}
(m+1)! = (m+1)m! > (m+1)^{10}
\end{align}
Also note that $(m+1)^{10} = (m+1)(m+1)^9$. So, cancelling the ''$m+1$'' terms, we need to show $m! > (m+1)^9$ using the fact that $m! > m^{10}$. 
Finally, notice that if we can show that $m^{10} > (m+1)^9$ then we are done. Equivalently, we just need to show that $\frac{m^{10}}{(m+1)^9} > 1$. I'll leave this up to you.
A: This is just a fun observation:
$$
(n!)^2= \prod_{i=1}^n i(n+1-i)>n^n
$$
so at least for $n\geq 20$ we have $\sqrt{(n!)^2}=n!>n^{n/2}\geq n^{10}$. Given this, we would only need to account for the cases $n=15,...,19$. But this is not an inductive approach, of course.
A: You need to work in the fact that $n! \gt n^{10}$ as that is the heart of the induction.  So for $n \gt 15$
$$\text {Base case } 15!-15^{10}=731023977375 \gt 0\\ \text {Assume }n! \gt n^{10}\\(n+1)! =(n+1)n!\gt (n+1)n^{10}=(n+1)^{10}\frac {n^{10}}{(n+1)^9}$$ Now we need to argue that the last fraction on the right is greater than $1$ and we are home.
$$\frac {n^{10}}{(n+1)^9} = n\left(1-\frac 1{n+1}\right)^9\gt n\left(1-\frac 9{n+1}\right)\gt 15\cdot \frac 7{16}\gt 1$$
A: Let $n\geq 15$ such that
$$n!\geq n^{10}$$
we want that
$$(n+1)!=(n+1)n!\geq(n+1)n^{10}$$
$$\geq (n+1)^{10}$$
which means that we want to prove 
$$n\geq (1+\frac 1n)^9$$
or
$$n^{\frac{n}{9}}\geq (1+\frac 1n)^n$$
and this is true for large enough $n$ cause the sequence on the left goes to $+\infty$, while the sequence on the right goes to $e$.
A: The handiest inequalities
for $n!$ are
$(n/e)^n < n! < (n/e)^{n+1}$.
These are easily proved
by induction from
$(1+1/n)^n < e < (1+1/n)^{n+1}$,
and these have been proved here
a number of times.
From
$n! > (n/e)^n$,
to show that
$n! > n^k$
for large enough $n$,
it is enough to show that
$(n/e)^n > n^k$
for large enough $n$.
$(n/e)^n > n^k
\iff
n^{n-k} > e^n
\iff
n^{1-k/n} > e
$.
If you are not worrying about
getting the best possible bound,
take $n > 2k$.
Then
$n^{1-k/n}
> n^{1/2}
> (2k)^{1/2} \gt e
$
for
$2k > e^2$.
If $2k < e^2$,
choose
$n > \max(2k, e^2)$.
Therefore,
if $n > \max(2k, e^2)$
then
$n! > n^k$.
A: There's no need for induction or anything beyond elementary arithmetic. For $n\geq 15$,
\begin{align*}
    n!&\geq n(n-1)\cdots(n-9)\times 5\cdot 4\cdot 3\cdot 2 \\
      &= 120n(n-1)\cdots(n-9)\\
      &> n \cdot \frac{15}{14}(n-1)\cdot \frac{15}{13}(n-2)\cdots \frac{15}{6}(n-9)\\
      &\geq n^{10}\,,
\end{align*}
where the strict inequality is because the third line is approximately $53n(n-1)\cdots(n-9)$ and the final inequality is because, for all $i$ and $n$ with $1\leq i<15\leq n$,
$$\frac{15}{15-i}(n-i) = n + \frac{(n-15)i}{15-i} \geq n\,.$$
A: Replacing $n$ by $n+1$, the LHS is multiplied by $n+1$ while the RHS is multiplied by $\left(1+\frac1n\right)^{10}$, which is bounded (by $1024$ for $n\ge1$, but by $2$ for $n\ge15$).
A: Since it's needed, here's the base case of $15! > 15^{10}$ without (much) calculation:
$15! = 2^{7}\cdot 2^{3}\cdot{2}\cdot 3^{5}\cdot{3}\cdot 5^{3}\cdot 7^2\cdot 11\cdot 13 \\ 
\phantom{15!} = 2^{11}\cdot 3^{6}\cdot 5^{3}\cdot 7^2\cdot 11\cdot 13 \\
\phantom{15!} =  16^2\cdot 15^3\cdot 18 \cdot (14\cdot 21)\cdot (22\cdot13) \\
\phantom{15!} > 15^2\cdot 15^3 \cdot 15\cdot 15^2\cdot 15^2 = 15^{10}$
