Throw the calculator away!

$\displaystyle \begin{aligned} 10^{100} & > k! & \small \color{#3D99F6} \text{Stirling's formula} \\ 10^{100} & > \sqrt{2 \pi k} \left( \frac ke \right) ^k & \small \color{#3D99F6} \text{Take the natural log of both sides} \\ 100 \ln 10 & > \ln \sqrt{2\pi} + \ln \sqrt{k} + k \ln k - k \ln e \\ 100 \ln 10 - \ln \sqrt{2\pi} & > \left(\frac 12 + k\right)\ln k - k \\ 229.34 & > \left(\frac 12 + k\right)\ln k - k \end{aligned}$

Eyeballing the equation above with $\ln k$ , we note that $e^4 < k < e^5$ :

$\displaystyle k = e^4 \Longrightarrow \left(\frac 12 + e^4\right) \cdot 4 - e^4 = 165.79 < 229.34 \\ k = e^5 \Longrightarrow \left(\frac 12 + e^5\right) \cdot 5 - e^5 = 596.15 > \ 229.34$

Clearly, $k$ must be closer to $e^4$ than to $e^5$ , so we make further approximations for $k$ . By testing some rational exponents of $e$ between $4$ and $5$ , we might find that $e^{4.25} = 70.1$ sets a very precise upper bound for $k$ :

$\displaystyle k = e^{4.25} \Longrightarrow \left(\frac 12 + e^{4.25}\right) \cdot 4.25 - e^{4.25} = 229.97 > 229.34$

Since $k$ is an integer, we test the integers $k \leq 70$ :

$\displaystyle \begin{aligned} k & = 70 \Longrightarrow \left(\frac 12 + 70\right) \ln 70 - 70 = 229.51 \ {\color{#D61F06}{\cancel{<}}} \ 229.34 \\ n & = 69 \Longrightarrow \left(\frac 12 + 69\right) \ln 69 - 69 = 225.27 \ {\color{#3D99F6}{<}} \ 229.34 \end{aligned}$

Hence, $k = \boxed{69}$ .

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Alternate solution

We could also find the number of digits in $k!$ :

$d(k!) = \big \lfloor \log_{10} k! \big \rfloor + 1$

We see that $k$ must be quite large for $\lfloor{k!\rfloor} = 10^{100}$ , so we can reason that $(k + 1)!$ must have more digits than $k!$ . Since $d\left(10^{100}\right) = 101$ and $10^{100}$ is the smallest number with $101$ digits, we have

$\displaystyle \begin{aligned} d(k!) < 101 \leq d\bigg((k + 1)!\bigg) \tag{1} \end{aligned}$

Stirling's formula says that for any positive integer $k$ , we have the bounds $\sqrt{2\pi}\ k^{\left(k+\frac 12\right)}e^{-k} \le k! \le e\ k^{\left(k+\frac 12\right)}e^{-k}$

so for $\left \lfloor \log_{10} k! \right \rfloor + 1$ ,

$\Bigg \lfloor \log_{10} \sqrt{2\pi} + \left(k + \frac 12\right) \log_{10} k - k \log_{10} e \Bigg \rfloor + 1 \ \leq \ \big \lfloor \log_{10} k! \big \rfloor + 1 \ \leq \ \Bigg \lfloor \log_{10} e + \left(k + \frac 12\right) \log_{10} k - k \log_{10} e \Bigg \rfloor + 1$

Combining this result with our previous inequality, $(1)$ , and doing some checking, we find that $69!$ has $99 < 101$ digits and $70!$ has $101 \leq 101$ digits.

Hence, $k = \boxed{69}$ .