Finding $\lim_{n\to \infty}(n!-10^n)$ $$\lim_{n\to \infty}(n!-10^n)$$
So of course $10^n<<n!$ but how can we prove that the limit is $\infty$?
 A: You can write the sequence as $n!\left(1-\frac{10^n}{n!}\right)$, then the result simply follows from arithmetic of limits. 
A: Hint:
For $m>10$,
$$(m+n)!>m!\,10^n.$$
Find $m$ such that $m!>10^m$.
A: Write $n! -10^n = 10^n \left(\frac {n!}{10^n} - 1 \right)$.
Now, $10^n \to \infty$. Let $b_n = \frac {n!}{10^n} - 1$. Then, $b_n+1 = \frac{n!}{10^n}$, and $b_{10}+1>0$ obviously.
Observe that $\frac{b_{n+1}+1}{b_n+1} = \frac{n+1}{10} \geq \frac{11}{10}$ for $n\geq 10$.
Therefore, $b_{n} \geq \left(\frac{11}{10}\right)^{n-10} \left(b_{10}+1\right)-1$ for $n \geq 11$. Thus, by domination, we see that $b_{n} \to +\infty$ as $n \to +\infty$.
Finally, $n!-10^n = 10^nb_n \to +\infty$.
A: Starting from @Yves Daoust's answer, you need to solve for $n$ the equation
$$n!=10^n$$
If you look at this question of mine, asking for the approximate solution of 
$$n!=a^n\, 10^k$$ you will find a magnificent approximation provided by @robjohn, an eminet user on this site.
Making $k=0$ and $a=10$, the answer is
$$n \sim 10\,e \exp{\left(W\left(-\frac{\log (20 \pi )}{20 e}\right) \right) }-\frac 12$$ where $W(.)$ is Lambert function.
Since the argument is quite small $\left(\frac{\log (20 \pi )}{20 e} \approx 0.076\right)$, you can use the series expansion
$$e^{W(x)}=1+x-\frac{1}{2}x^2+\frac{2 }{3}x^3+O\left(x^4\right)$$ to get $n=24.5257$ then $n=25$.
Numerically, the exact solution of the equation is $24.5264$ (!!)
A: Yet another approach, inspired by the one of Yves Daoust. Assume $n > 100$ and write $n=100+m$. Then $n! > 100^m = 10^{2m}$. So in that case $$n! - 10^n > 10^{2m} - 10^{100+m} = 10^m \cdot (10^m - 10^{100}).$$ Now as we let $m$ tend to infinity, we observe that both terms of the product tend to infinity. 
A: $$e^{10}= 1+10/1!+10^2/2!+\cdots + 10^k/k!+ \cdots$$
Since this series is convergent, it follows that $\lim_{k \rightarrow \infty}10^k/k!=0$.
Hence, for sufficiently large $k:$ $10^k/k! <1/2$.
$k!(1-10^k/k!)>k!(1-1/2)=(1/2)k!$.
Another attempt:
Note: $k!>(k/2)^{k/2}$; for $k >2$.
$f(k):= k!(1-10^k/k!)$;
$f(k) >k!(1-\dfrac{100^{k/2}}{(k/2)^{k/2}})=$
$k!(1-(\dfrac{100}{k/2})^{k/2})$.
Take the limit.
