A simple upper bound:

$7^2 = 49$ is just under $50$, so $7^{100} = (7^2)^{50}$ is somewhat less than $50^{50} = (100/2)^{50}$.

$2^{10} = 1024$ is just over $1000 = 10^3$, so $2^{50}$ is somewhat over $10^{15}$.

So $7^{50} < (100^{50}) / (10^{15}) = ((10^2)^{50}) / (10^{15}) = (10^{100}) / (10^{15}) = 10^{85} = 1\mathrm{e}85$.

And a semi-simple lower bound:

$7^6 = 117649$ is somewhat more than $10^5$, so $7^{96} = (7^6)^{16}$ is somewhat more than $(10^5)^{16} = 10^{80}$.

$7^{100} = (7^{96}) * (7^4)$ is somewhat more than $10^{80} * 2401 = 2.40 1\mathrm{e}83$.

Or, borrowing an answer from the previous problem: If you have

a log table giving $log_{10}(7) \approx 0.845$

then you can determine that

$log_{10}(7^{100}) = 100 * log_{10}(7) \approx 84.5$

which produces an estimate of

$\sqrt{10} * 10^{84} ~= 3.162\mathrm{e}84$

A simple upper bound:

$7^2 = 49$ is just under $50$, so $7^{100} = (7^2)^{50}$ is somewhat less than $50^{50} = (100/2)^{50}$.

$2^{10} = 1024$ is just over $1000 = 10^3$, so $2^{50}$ is somewhat over $10^{15}$.

So $$7^{50} < \frac{100^{50}}{10^{15}} = \frac{\left(10^2\right)^{50}}{10^{15}} = \frac{10^{100}}{10^{15}} = 10^{85} = 1\ \text{e}\ 85$$

And a semi-simple lower bound:

$7^6 = 117649$ is somewhat more than $10^5$, so $7^{96} = \left(7^6\right)^{16}$ is somewhat more than $\left(10^5\right)^{16} = 10^{80}$.

$7^{100} = 7^{96} \times 7^4$ is somewhat more than $10^{80} \times 2401 = 2.40 1\ \mathrm{e}\ 83$.

Or, borrowing an answer from the previous problem: If you have

a log table giving $\log_{10}(7) \approx 0.845$

then you can determine that

$\log_{10}(7^{100}) = 100 \times \log_{10}(7) \approx 84.5$

which produces an estimate of

$\sqrt{10} \times 10^{84} \approx 3.162\ \mathrm{e}\ 84$

A simple upper bound:

7^2 = 49 is just under 50, so 7^100 = (7^2)^50 is somewhat less than 50^50 = (100/2)^50.

2^10 = 1024 is just over 1000 = 10^3, so 2^50 is somewhat over 10^15.

So 7^50 is less than (100^50) / (10^15) = ((10^2)^50) / (10^15) = (10^100) / (10^15) = 10^85 (1E85).

And a semi-simple lower bound:

7^6 = 117649 is somewhat more than 10^5, so 7^96 = (7^6)^16 is somewhat more than (10^5)^16 = 10^80.

7^100 = (7^96) * (7^4) is somewhat more than 10^80 * 2401 = 2.401E83.

Or, borrowing an answer from the previous problem: If you have

a log table giving log10(7) ~= 0.845

then you can determine that

log10(7^100) = 100 * log10(7) ~= 84.5

which produces an estimate of

sqrt(10) * 10^84 ~= 3.162E84

A simple upper bound:

$7^2 = 49$ is just under $50$, so $7^{100} = (7^2)^{50}$ is somewhat less than $50^{50} = (100/2)^{50}$.

$2^{10} = 1024$ is just over $1000 = 10^3$, so $2^{50}$ is somewhat over $10^{15}$.

So $7^{50} < (100^{50}) / (10^{15}) = ((10^2)^{50}) / (10^{15}) = (10^{100}) / (10^{15}) = 10^{85} = 1\mathrm{e}85$.

And a semi-simple lower bound:

$7^6 = 117649$ is somewhat more than $10^5$, so $7^{96} = (7^6)^{16}$ is somewhat more than $(10^5)^{16} = 10^{80}$.

$7^{100} = (7^{96}) * (7^4)$ is somewhat more than $10^{80} * 2401 = 2.40 1\mathrm{e}83$.

Or, borrowing an answer from the previous problem: If you have

a log table giving $log_{10}(7) \approx 0.845$

then you can determine that

$log_{10}(7^{100}) = 100 * log_{10}(7) \approx 84.5$

which produces an estimate of

$\sqrt{10} * 10^{84} ~= 3.162\mathrm{e}84$