Euler found out that Euler's number was the basis of the natural logarithm. He started with this equation for infinitely small numbers $w$:
For $a>1$:
$$a^w=1+kw$$ $w$=infinitely small numbers
$a$=base
From $a^w=1+kw$ to this: $${a}^{jw}={(1+kw)}^{j}$$ $${a}^{jw}={(1+kw)}^{j}=1+\frac{j}{1}kw+\frac{j(j-1)}{1*2}{k}^{2}{w}^{2}+\frac{j(j-1)(j-2)}{1*2*3}{k}^{3}{w}^{3}+...$$ Then he replaced $j$ with $\frac{z}{w}$: $$j=\frac{z}{w}$$ $z$ is any finite number, $ω$ is an infinitely small number and $j$ is an infinitely large number.
Then he replaced $w$ with $\frac{z}{j}$: $${a}^{z}={a}^{jw}={(1+k*\frac{z}{j})}^{j}=1+\frac{j}{1}k*\frac{z}{j}+\frac{j(j-1)}{1*2}{k}^{2}{(\frac{z}{j})}^{2}+\frac{j(j-1)(j-2)}{1*2*3}{k}^{3}{(\frac{z}{j})}^{3}+...$$ Since $j$ is infinitely large, you will get this: $${a}^{z}=1+\frac{kz}{1}+\frac{{k}^{2}{z}^{2}}{1*2}+\frac{{k}^{3}{z}^{3}}{1*2*3}+...$$
Then you replace $z$ and $k$ with $1$ to get the base of natural logarithm: $${a}=1+\frac{1}{1}+\frac{1}{1*2}+\frac{1}{1*2*3}+...$$ My question is, why did Euler replace $k$ and $z$ with $1$ to get the base of natural logarithm? He had made also a series for $$\log{1+x}=\frac{1}{k}(\frac{x}{1}-\frac{{x}^{2}}{2}+\frac{{x}^{3}}{3}-\frac{{x}^{4}}{4}+...)$$ You can see that here in this link from page 23 to 25. Here is the link
I don't claim to really understand Euler's calculation, but here's my take on it. Obviously, to compute $a$ for $a^z,$ you set $z=1,$ so the only question is why Euler sets $k=1.$ The equation $$a^w=1+kw$$ for infinitesimal $w$ means, in standard terminology, $$k= {da^w\over dw}|_{w=0}=\ln a$$ so if $a$ is the base of natural logarithms, $k=1.$
EDIT
What I mean to say is that $a^w=1+kw$ is the same as $$k={a^w-1\over w}$$ for infinitesimal $w$ and this means that $k$ is the derivative of $a^w$ (as a function of $w$) at $w=0$. Now if $a=e,$ we must have $k=1$. I haven't read Euler's calculation, but you say that he was determining the base for natural logarithms, and I take that to mean the number $a$ such that $\ln a = 1.$