Showing n! is greater than n to the tenth power

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.

$$\frac {n^{10}}{(n+1)^9} = n\left(1-\frac 1{n+1}\right)^9\gt n\left(1-\frac 9{n+1}\right)\gt 15\cdot \frac 7{16}\gt 1$$

Since it's needed, here's the base case of $15! > 15^{10}$ without (much) calculation:

$15! = 2^{7}\cdot 2^{3}\cdot{2}\cdot 3^{5}\cdot{3}\cdot 5^{3}\cdot 7^2\cdot 11\cdot 13 \\ \phantom{15!} = 2^{11}\cdot 3^{6}\cdot 5^{3}\cdot 7^2\cdot 11\cdot 13 \\ \phantom{15!} = 16^2\cdot 15^3\cdot 18 \cdot (14\cdot 21)\cdot (22\cdot13) \\ \phantom{15!} > 15^2\cdot 15^3 \cdot 15\cdot 15^2\cdot 15^2 = 15^{10}$

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.