If you the objective allows all calculations to be performed in Mathematica (rather than reproduced in Excel), then using the Interpolation function works great:

data = {{0., 0.}, {0.05, 0.4651}, {0.1, 0.6433}, {0.15, 0.7298}, {0.2,
     0.779}, {0.25, 0.81}, {0.3, 0.831}, {0.35, 0.8462}, {0.4, 
    0.8576}, {0.45, 0.8664}, {0.5, 0.8736}, {0.55, 0.8796}, {0.6, 
    0.8848}, {0.65, 0.8895}, {0.7, 0.8941}, {0.75, 0.8989}, {0.8, 
    0.9045}, {0.85, 0.9118}, {0.9, 0.9231}, {0.95, 0.9445}, {1., 1.}};
f = Interpolation[data]
Show[{ListPlot[data], Plot[f[x], {x, 0, 1}, PlotRange -> All]}]
f[0.756]

with output

But if you need the fitted equation outside of Mathematica (like in Excel), then determining how Mathematica deals with the interpolation object might be possible but I don't know if that's available. If providing predictions to others is the objective, does turning this into a CDF document achieve that?

Excellent about the use of StringReplace! I'm embarrassed to say that all these years I've been editing those output equations by hand. Thank you!

I wouldn't necessarily recommend polynomials at all. I should have asked initially how your data was generated. As a statistician, I'm always suspicious of data with such a smooth connected set of points. Is this one run of some experiment? Are the points generated by a much more complex formula and you're trying to approximate that with something that faster to run? Is the process that generates the data inducing a serial correlation? In other words, what is the data generation process? (I ask because LinearModelFit as used assumes that the errors are independent and identically distributed - which is important if you want to make inferences about the parameters or predictions.)

There are two issues:

1. Unless this is for an unrealistic homework problem, you should never try to fit 14 parameters to 20 data points.
2. But to really answer your question, you didn't obtain enough significant digits to describe accurately enough the curve (i.e., round-off error did you in). While this kind of thing can happen even with just a few parameters, it's much more likely to happen when you attempt to overfit.

Below is some Mathematica code that gives almost the desired form to put into Excel:

FortranForm[Normal[lm]]

You'll get the following:

     4.5287284920657044e-7 + 15.253009866430183*x - 172.24927379188136*x**2 + 1400.1849477287135*x**3 - 8217.174024928278*x**4 + 34923.989035280516*x**5 - 
     -  107985.6698971923*x**6 + 243429.10048833332*x**7 - 398597.8815182028*x**8 + 467992.1961978472*x**9 - 383346.5315456258*x**10 + 207812.90647534237*x**11 - 
     -  66956.90233584394*x**12 + 9703.778441043793*x**13

You'll need to add an "=" to the beginning of the first row, remove the two "-" from the beginning of the second and third lines, change the "**" to "^", and change "x" to a Excel cell address something like the following:

=4.5287284920657E-07 + 15.2530098664301*A2 - 172.249273791881*A2^2 + 1400.18494772871*A2^3 - 8217.17402492827*A2^4 + 34923.9890352805*A2^5 - 107985.669897192*A2^6 + 243429.100488333*A2^7 - 398597.881518202*A2^8 + 467992.196197847*A2^9 - 383346.531545625*A2^10 + 207812.906475342*A2^11 -66956.9023358439*A2^12 + 9703.77844104379*A2^13

I've attached an Excel workbook with the resulting data and associated curve.

Attachments: