Lebesgue measure using special polygons

I want to prove that the following set satisfies $m(G)=1$.

$G=\{(x,y)\in {\mathbb{R}}^{2}:x>0,0<y<{e}^{-x}\}$

I have to prove this without using the Lebesgue integral.

This set is open, so its measure is defined as it follows:

$m(G)=sup\{m(P):P\subset G\}$, where P is a special polygon.

We can construct a sequence of special polygons ${P}_{n}\subset G$:

${P}_{n}=\bigcup _{j=1}^{n}[0,log\left(\frac{n+1}{j}\right)]\times [\frac{j-1}{n+1},\frac{j}{n+1}].$

The measure of one of these special polygons is $m({P}_{n})=\sum _{j=1}^{n}log\left(\frac{n+1}{j}\right)\frac{1}{n+1}.$

As the sequence of measures $\{m({P}_{n})\}$ is increasing, the supremum of this sequence is its limit:

$sup(\{m({P}_{n})\})=\underset{n\to \mathrm{\infty}}{lim}\sum _{j=1}^{n}log\left(\frac{n+1}{j}\right)\frac{1}{n+1}=\underset{n\to \mathrm{\infty}}{lim}\mathrm{log}{\left(\frac{(n+1{)}^{n}}{n!}\right)}^{\frac{1}{n+1}}.$

I know that this limit is 1, so the logarithm argument must tend to e, but I don't know how to prove this analytically. I've tried working on the final expression to get the definition of e, but I haven't been able to do so. Could yo help me?

I want to prove that the following set satisfies $m(G)=1$.

$G=\{(x,y)\in {\mathbb{R}}^{2}:x>0,0<y<{e}^{-x}\}$

I have to prove this without using the Lebesgue integral.

This set is open, so its measure is defined as it follows:

$m(G)=sup\{m(P):P\subset G\}$, where P is a special polygon.

We can construct a sequence of special polygons ${P}_{n}\subset G$:

${P}_{n}=\bigcup _{j=1}^{n}[0,log\left(\frac{n+1}{j}\right)]\times [\frac{j-1}{n+1},\frac{j}{n+1}].$

The measure of one of these special polygons is $m({P}_{n})=\sum _{j=1}^{n}log\left(\frac{n+1}{j}\right)\frac{1}{n+1}.$

As the sequence of measures $\{m({P}_{n})\}$ is increasing, the supremum of this sequence is its limit:

$sup(\{m({P}_{n})\})=\underset{n\to \mathrm{\infty}}{lim}\sum _{j=1}^{n}log\left(\frac{n+1}{j}\right)\frac{1}{n+1}=\underset{n\to \mathrm{\infty}}{lim}\mathrm{log}{\left(\frac{(n+1{)}^{n}}{n!}\right)}^{\frac{1}{n+1}}.$

I know that this limit is 1, so the logarithm argument must tend to e, but I don't know how to prove this analytically. I've tried working on the final expression to get the definition of e, but I haven't been able to do so. Could yo help me?