Approximation of $26!$ Peltzl's Cryptology states on page 8 that $26!$ is approximately $2^{88}$. I have tried different variations of Stirling's formula to confirm this but no luck. I know the argument is hiding in there somewhere and a countable infinity of heads are better than one. Any help out there?
 A: Write it as
$$26!=2^{23}\cdot3^{10}\cdot5^6\cdot7^3\cdot11^2\cdot13^2\cdot17\cdot19\cdot23\;,$$
and take the log base $2$:
$$23+10\lg3+6\lg5+3\lg7+2\lg11+2\lg13+\lg17+\lg19+\lg23$$
is approximately $88.382$, so $88$ is the nearest integer exponent.
A: As suggested by vadim123:
$$\log_2(26!) = \sum_{n=2}^{26}\log_2n$$
Approximating brutally with $\log_2n \approx \lfloor\log_2n\rfloor$ we get:
$$\log_2(26!)\approx 2\cdot1+4\cdot2+8\cdot3+11\cdot4=78$$
Thus with this (extremely inefficient) approximation:
$$26!\approx2^{78}$$
Summing the "exact" logarithms with Wolfram Alpha, we obtain
$$\log_2(26!)\approx88.4$$
$$\Rightarrow 26!\approx2^{88}$$
A: $
\ln\left(26!\right)
\sim
\left(26 + 1/2\right)\ln\left(26\right) - 26 + \ln\left(2\pi\right)/2
\sim
n\ln\left(2\right)$
$$
n \sim {26.5\ln\left(26\right) -26 + \ln\left(2\pi\right)/2\over \ln\left(2\right)}
=
88.3773\ldots
$$
$$
\color{#ff0000}{\large\ln\left(26!\right)}
\sim
2^{0.3373} \times 2^{88} \sim \color{#ff0000}{\large 1.2633 \times 2^{\color{#0000ff}{88}}}
$$
A: Stirling's Formula works:
$$ n! \sim (\frac{n}{e})^n\sqrt{2\pi n}$$
and so
$$\log_2{n!}=\frac{\ln{n!}}{\ln2} \sim \frac{n(\ln n-1)}{\ln 2}+\frac{\ln \pi +\ln n}{2 \ln 2}+\frac{1}{2}$$
This gives the approximation 
$$\log_2 26! \sim 88.377$$
for 
$$\log_2 26! = 88.381$$
